$\mathrm{restart}$

$\mathrm{with}\left(\mathrm{Student}\left[\mathrm{Calculus1}\right]\right)\:$

$\mathrm{infolevel}\left[\mathrm{Student}\left[\mathrm{Calculus1}\right]\right]≔1\:$

$\underset{x\→3}{lim}\left(x^2\right)$

$\underset{x\to 3}{lim}{x}^2$

$\mathrm{Rule}\left[\mathrm{power}\right]\left(\right)$

Creating problem #1

$\underset{x\to 3}{lim}{x}^2={\left(\underset{x\to 3}{lim}x\right)}^2$

$\mathrm{Rule}\left[\mathrm{identity}\right]\left(\right)$

$\underset{x\to 3}{lim}{x}^2=9$

$\mathrm{ShowSteps}\left(\right)$

$\begin{array}{cccc}\multicolumn{4}{c}{\underset{x\to 3}{lim}{x}^2}\\ & \text{=}& {\left(\underset{x\to 3}{lim}x\right)}^2\hfill & \left[\mathrm{power}\right]\\ & \text{=}& 9\hfill & \left[\mathrm{identity}\right]\end{array}$

$\mathrm{Rule}\left[\mathrm{product}\right]\left(\frac{\ⅆ}{\ⅆx}\left(x^2\sin \left(x\right)\right)\right)$

Creating problem #2

$\frac{\ⅆ}{\ⅆx}\left(x^2\sin \left(x\right)\right)=\left(\frac{\ⅆ}{\ⅆx}\left(x^2\right)\right)\sin \left(x\right)+x^2\left(\frac{\ⅆ}{\ⅆx}\sin \left(x\right)\right)$

$\mathrm{Rule}\left[\mathrm{`^`}\right]\left(\right)$

$\frac{\ⅆ}{\ⅆx}\left(x^2\sin \left(x\right)\right)=2x\sin \left(x\right)+x^2\left(\frac{\ⅆ}{\ⅆx}\sin \left(x\right)\right)$

$\mathrm{Rule}\left[\sin \right]\left(\right)$

$\frac{\ⅆ}{\ⅆx}\left(x^2\sin \left(x\right)\right)=2x\sin \left(x\right)+x^2\cos \left(x\right)$

$\mathrm{GetProblem}\left(1\right)$

$\underset{x\to 3}{lim}{x}^2=9$

$\mathrm{Show}\left(2\right)$

$\frac{\ⅆ}{\ⅆx}\left(x^2\sin \left(x\right)\right)=2x\sin \left(x\right)+x^2\cos \left(x\right)$

$\mathrm{Show}\left(\mathrm{all}\right)$

Problem 1:

$\underset{x\to 3}{lim}{x}^2=9$

Problem 2:

$\frac{\ⅆ}{\ⅆx}\left(x^2\sin \left(x\right)\right)=2x\sin \left(x\right)+x^2\cos \left(x\right)$

$\frac{\ⅆ}{\ⅆx}\sin \left(x^2\right)$

$\frac{\ⅆ}{\ⅆx}\sin \left(x^2\right)$

$\mathrm{Hint}\left(\right)$

Creating problem #3

$\left[\mathrm{chain}\right]$

$\mathrm{Rule}\left[\right]\left(\right)$

$\frac{\ⅆ}{\ⅆx}\sin \left(x^2\right)=\left(\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆ\mathrm{\_X0}}\sin \left(\mathrm{\_X0}\right)\right)}{\phantom{\mathrm{\_X0}=x^2}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆ\mathrm{\_X0}}\sin \left(\mathrm{\_X0}\right)\right)}}{\mathrm{\_X0}=x^2}\right)\left(\frac{\ⅆ}{\ⅆx}\left(x^2\right)\right)$

$\∫x\cos \left(x\right)\ⅆx$

$\int x\cos \left(x\right)\ⅆx$

$\mathrm{Hint}\left(\right)$

Creating problem #4

$\left[\mathrm{parts}\,x\,\sin \left(x\right)\,1\,\cos \left(x\right)\,x\right]$

$\mathrm{Rule}\left[\right]\left(\right)$

$\int x\cos \left(x\right)\ⅆx=x\sin \left(x\right)-\left(\int \sin \left(x\right)\ⅆx\right)$

$\underset{x\→0^{\+}}{lim}{x}^x$

$\underset{x\to 0+}{lim}{x}^x$

$\mathrm{Hint}\left(\right)$

Creating problem #5

Rewrite the expression as an exponential to prepare for using l'Hôpital's rule

$\left[\mathrm{rewrite}\,x^x={\ⅇ}^{x\ln \left(x\right)}\right]$

$\mathrm{Rule}\left[\right]\left(\right)$

$\underset{x\to 0+}{lim}{x}^x=\underset{x\to 0+}{lim}{\ⅇ}^{x\ln \left(x\right)}$

$\mathrm{Rule}\left[\exp \right]\left(\right)$

$\underset{x\to 0+}{lim}{x}^x={\ⅇ}^{\underset{x\to 0+}{lim}x\ln \left(x\right)}$

$\mathrm{Hint}\left(\right)$

$\left[\mathrm{lhopital}\,\ln \left(x\right)\right],\left[\mathrm{lhopital}\,x\right]$

$\mathrm{Rule}\left[\right]\left(\right)$

$\underset{x\to 0+}{lim}{x}^x={\ⅇ}^{\underset{x\to 0+}{lim}\left(-x\right)}$

$\mathrm{Rule}\left[\mathrm{`c*`}\right]\left(\right)$

$\underset{x\to 0+}{lim}{x}^x={\ⅇ}^{-\left(\underset{x\to 0+}{lim}x\right)}$

$\mathrm{Rule}\left[\mathrm{identity}\right]\left(\right)$

$\underset{x\to 0+}{lim}{x}^x=1$

$\mathrm{Understand}\left(\mathrm{Diff}\,\mathrm{constant}\,\mathrm{c*}\,\mathrm{`+`}\right)$

$\mathrm{Diff}=\left[\mathrm{constant}\,\mathrm{constantmultiple}\,\mathrm{sum}\right]$

$\mathrm{Rule}\left[\mathrm{`*`}\right]\left(\frac{\ⅆ}{\ⅆx}\left(x^2\left(\cos \left(x\right)-2\ln \left(x\right)\right)\right)\right)$

Creating problem #6

$\frac{\ⅆ}{\ⅆx}\left(x^2\left(\cos \left(x\right)-2\ln \left(x\right)\right)\right)=\left(\frac{\ⅆ}{\ⅆx}\left(x^2\right)\right)\left(\cos \left(x\right)-2\ln \left(x\right)\right)+x^2\left(\frac{\ⅆ}{\ⅆx}\cos \left(x\right)-2\frac{\ⅆ}{\ⅆx}\ln \left(x\right)\right)$

${\mathrm{Rule}}_{}\left[\right]\left(\frac{\ⅆ}{\ⅆx}\left(3x\+{\ⅇ}^x\right)\right)$

Creating problem #7

$\frac{\ⅆ}{\ⅆx}\left(3x+{\ⅇ}^x\right)=3\frac{\ⅆ}{\ⅆx}x+\frac{\ⅆ}{\ⅆx}{\ⅇ}^x$

${\∫}_a^b\frac{w^{\frac{1}{3}}}{w^{\frac{1}{2}}\+1}\ⅆw$

${\int }_a^b\frac{w^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{\sqrt{w}+1}\ⅆw$

$\mathrm{Hint}\left(\right)$

Creating problem #8

$\left[\mathrm{change}\,w=u^6\,u\right]$

$\mathrm{Rule}\left[\right]\left(\right)$

Applying substitution w = u^6, u = w^(1/6) with dw = 6*u^5*du, du = 1/6/w^(5/6)*dw

${\int }_a^b\frac{w^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{\sqrt{w}+1}\ⅆw={\int }_{w=a}^{w=b}\left(6u^4-6u+\frac{6u}{u^3+1}\right)\ⅆu$

$\mathrm{Understand}\left(\mathrm{int}\,\mathrm{constant}\,\mathrm{c*}\,\mathrm{`+`}\right)$

$\mathrm{Int}=\left[\mathrm{constant}\,\mathrm{constantmultiple}\,\mathrm{sum}\right]$

$\mathrm{Rule}\left[\right]\left(\right)$

${\int }_a^b\frac{w^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{\sqrt{w}+1}\ⅆw=6\left({\int }_{w=a}^{w=b}{u}^4\ⅆu\right)-6\left({\int }_{w=a}^{w=b}u\ⅆu\right)+6\left({\int }_{w=a}^{w=b}\frac{u}{u^3+1}\ⅆu\right)$

$\mathrm{Rule}\left[\mathrm{Hint}\left(\right)\right]\left(\right)$

${\int }_a^b\frac{w^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{\sqrt{w}+1}\ⅆw=6\left(\genfrac{}{}{0ex}{}{\frac{u^5}{5}}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\frac{u^5}{5}}}{w=a..b}\right)-6\left({\int }_{w=a}^{w=b}u\ⅆu\right)+6\left({\int }_{w=a}^{w=b}\frac{u}{u^3+1}\ⅆu\right)$

$\mathrm{Rule}\left[\mathrm{Hint}\left(\right)\right]\left(\right)$

${\int }_a^b\frac{w^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{\sqrt{w}+1}\ⅆw=6\left(\genfrac{}{}{0ex}{}{\frac{u^5}{5}}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\frac{u^5}{5}}}{w=a..b}\right)-6\left(\genfrac{}{}{0ex}{}{\frac{u^2}{2}}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\frac{u^2}{2}}}{w=a..b}\right)+6\left({\int }_{w=a}^{w=b}\frac{u}{u^3+1}\ⅆu\right)$

$\mathrm{Hint}\left(\right)$

$\left[\mathrm{partialfractions}\right]$

$\mathrm{Rule}\left[\right]\left(\right)$

${\int }_a^b\frac{w^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{\sqrt{w}+1}\ⅆw=6\left(\genfrac{}{}{0ex}{}{\frac{u^5}{5}}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\frac{u^5}{5}}}{w=a..b}\right)-6\left(\genfrac{}{}{0ex}{}{\frac{u^2}{2}}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\frac{u^2}{2}}}{w=a..b}\right)+2\left({\int }_{w=a}^{w=b}\frac{u+1}{u^2-u+1}\ⅆu\right)-2\left({\int }_{w=a}^{w=b}\frac{1}{u+1}\ⅆu\right)$

$\mathrm{ShowIncomplete}\left(\right)$

$\mathrm{Int23}={\int }_{w=a}^{w=b}\frac{u+1}{u^2-u+1}\ⅆu$

$\mathrm{Int24}={\int }_{w=a}^{w=b}\frac{1}{u+1}\ⅆu$

$L:=\mathrm{ShowIncomplete}\left(\mathrm{data}\right)\&colon;$$\mathrm{lbl}:=\mathrm{if}\left(\mathrm{length}\left(L_2\right)<\mathrm{length}\left(L_1\right)\&comma;L_{1\&comma;1}\&comma;L_{2\&comma;1}\right)$

$\mathrm{lbl}≔\mathrm{Int23}$

$\mathrm{Hint}\left(\mathrm{lbl}\right)$

Rewrite the numerator in a form which contains the derivative of the denominator

$\left[\mathrm{rewrite}\&comma;\frac{u+1}{u^2-u+1}=\frac{2u-1}{2\left(u^2-u+1\right)}+\frac{3}{2\left(u^2-u+1\right)}\right]$

$\mathrm{Rule}\left[\right]\left(\mathrm{lbl}\right)$

${\int }_{w=a}^{w=b}\frac{u+1}{u^2-u+1}\&DifferentialD;u=\frac{\left({\int }_{w=a}^{w=b}\frac{2u-1}{u^2-u+1}\&DifferentialD;u\right)}{2}+\frac{3\left({\int }_{w=a}^{w=b}\frac{1}{u^2-u+1}\&DifferentialD;u\right)}{2}$

$\mathrm{Hint}\left(\right)$

Note that the derivative of u^2-u+1 is 2*u-1, so we can make a change of variable.

$\left[\mathrm{change}\&comma;v=u^2-u+1\&comma;v\right]$

$\mathrm{Rule}\left[\right]\left(\right)$

Applying substitution u = 1/2+1/2*(-3+4*v)^(1/2), v = u^2-u+1 with du = 1/(-3+4*v)^(1/2)*dv, dv = (2*u-1)*du

${\int }_{w=a}^{w=b}\frac{u+1}{u^2-u+1}\&DifferentialD;u=\frac{\left({\int }_{w=a}^{w=b}\frac{1}{v}\&DifferentialD;v\right)}{2}+\frac{3\left({\int }_{w=a}^{w=b}\frac{1}{u^2-u+1}\&DifferentialD;u\right)}{2}$

$\mathrm{Rule}\left[\mathrm{Hint}\left(\right)\right]\left(\right)$

${\int }_{w=a}^{w=b}\frac{u+1}{u^2-u+1}\&DifferentialD;u=\frac{\left(\genfrac{}{}{0ex}{}{\ln \left(v\right)}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\ln \left(v\right)}}{w=a..b}\right)}{2}+\frac{3\left({\int }_{w=a}^{w=b}\frac{1}{u^2-u+1}\&DifferentialD;u\right)}{2}$

$\mathrm{Hint}\left(\right)$

$\left[\mathrm{revert}\right]$

$\mathrm{Rule}\left[\right]\left(\right)$

Reverting substitution using v = u^2-u+1

${\int }_{w=a}^{w=b}\frac{u+1}{u^2-u+1}\&DifferentialD;u=\frac{\left(\genfrac{}{}{0ex}{}{\ln \left(u^2-u+1\right)}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\ln \left(u^2-u+1\right)}}{w=a..b}\right)}{2}+\frac{3\left({\int }_{w=a}^{w=b}\frac{1}{u^2-u+1}\&DifferentialD;u\right)}{2}$

$\mathrm{Hint}\left(\right)$

Complete the square and make a change of variable.

$\left[\mathrm{change}\&comma;v=u-\frac{1}{2}\&comma;v\right]$

$\mathrm{Rule}\left[\right]\left(\right)$

Applying substitution u = 1/2+v with du=dv

${\int }_{w=a}^{w=b}\frac{u+1}{u^2-u+1}\&DifferentialD;u=\frac{\left(\genfrac{}{}{0ex}{}{\ln \left(u^2-u+1\right)}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\ln \left(u^2-u+1\right)}}{w=a..b}\right)}{2}+6\left({\int }_{w=a}^{w=b}\frac{1}{4v^2+3}\&DifferentialD;v\right)$

$\mathrm{Rule}\left[\mathrm{Hint}\left(\right)\right]\left(\right)$

Integrals involving expressions of the form (x^2+a^2)^n or sqrt(x^2+a^2)^n can often be simplified using the substitution x = a*tan(u).
Applying substitution v = 1/2*3^(1/2)*tan(z), z = arctan(2/3*v*3^(1/2)) with dv = 1/2*3^(1/2)*(1+tan(z)^2)*dz, dz = 2/3*3^(1/2)/(4/3*v^2+1)*dv

${\int }_{w=a}^{w=b}\frac{u+1}{u^2-u+1}\&DifferentialD;u=\frac{\left(\genfrac{}{}{0ex}{}{\ln \left(u^2-u+1\right)}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\ln \left(u^2-u+1\right)}}{w=a..b}\right)}{2}+6\left(\genfrac{}{}{0ex}{}{\frac{\sqrt{3}z}{6}}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\frac{\sqrt{3}z}{6}}}{w=a..b}\right)$

$\mathrm{Rule}\left[\mathrm{revert}\right]\left(\right)$

Reverting substitution using z = arctan(2/3*v*3^(1/2))

${\int }_{w=a}^{w=b}\frac{u+1}{u^2-u+1}\&DifferentialD;u=\frac{\left(\genfrac{}{}{0ex}{}{\ln \left(u^2-u+1\right)}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\ln \left(u^2-u+1\right)}}{w=a..b}\right)}{2}+6\left(\genfrac{}{}{0ex}{}{\frac{\sqrt{3}\arctan \left(\frac{2v\sqrt{3}}{3}\right)}{6}}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\frac{\sqrt{3}\arctan \left(\frac{2v\sqrt{3}}{3}\right)}{6}}}{w=a..b}\right)$

$\mathrm{Rule}\left[\mathrm{revert}\right]\left(\right)$

Reverting substitution using v = u-1/2

${\int }_{w=a}^{w=b}\left(\frac{u+1}{3\left(u^2-u+1\right)}-\frac{1}{3\left(u+1\right)}\right)\&DifferentialD;u=\frac{\left(\genfrac{}{}{0ex}{}{\ln \left(u^2-u+1\right)}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\ln \left(u^2-u+1\right)}}{w=a..b}\right)}{6}+2\left(\genfrac{}{}{0ex}{}{\frac{\sqrt{3}\arctan \left(\frac{2\left(u-\frac{1}{2}\right)\sqrt{3}}{3}\right)}{6}}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\frac{\sqrt{3}\arctan \left(\frac{2\left(u-\frac{1}{2}\right)\sqrt{3}}{3}\right)}{6}}}{w=a..b}\right)-\frac{\left({\int }_{w=a}^{w=b}\frac{1}{u+1}\&DifferentialD;u\right)}{3}$

$\mathrm{Rule}\left[\mathrm{change}\&comma;\mathrm{u1}\&equals;u\&plus;1\right]\left(\right)$

Applying substitution u = -1+u1 with du=du1

${\int }_{w=a}^{w=b}\left(\frac{u+1}{3\left(u^2-u+1\right)}-\frac{1}{3\left(u+1\right)}\right)\&DifferentialD;u=\frac{\left(\genfrac{}{}{0ex}{}{\ln \left(u^2-u+1\right)}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\ln \left(u^2-u+1\right)}}{w=a..b}\right)}{6}+2\left(\genfrac{}{}{0ex}{}{\frac{\sqrt{3}\arctan \left(\frac{2\left(u-\frac{1}{2}\right)\sqrt{3}}{3}\right)}{6}}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\frac{\sqrt{3}\arctan \left(\frac{2\left(u-\frac{1}{2}\right)\sqrt{3}}{3}\right)}{6}}}{w=a..b}\right)-\frac{\left({\int }_{w=a}^{w=b}\frac{1}{\mathrm{u1}}\&DifferentialD;\mathrm{u1}\right)}{3}$

$\mathrm{Rule}\left[\mathrm{`^`}\right]\left(\right)$

${\int }_{w=a}^{w=b}\left(\frac{u+1}{3\left(u^2-u+1\right)}-\frac{1}{3\left(u+1\right)}\right)\&DifferentialD;u=\frac{\left(\genfrac{}{}{0ex}{}{\ln \left(u^2-u+1\right)}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\ln \left(u^2-u+1\right)}}{w=a..b}\right)}{6}+2\left(\genfrac{}{}{0ex}{}{\frac{\sqrt{3}\arctan \left(\frac{2\left(u-\frac{1}{2}\right)\sqrt{3}}{3}\right)}{6}}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\frac{\sqrt{3}\arctan \left(\frac{2\left(u-\frac{1}{2}\right)\sqrt{3}}{3}\right)}{6}}}{w=a..b}\right)-\frac{\left(\genfrac{}{}{0ex}{}{\ln \left(\mathrm{u1}\right)}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\ln \left(\mathrm{u1}\right)}}{w=a..b}\right)}{3}$

$\mathrm{Rule}\left[\mathrm{revert}\right]\left(\right)$

Reverting substitution using u1 = u+1

${\int }_a^b\frac{w^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{\sqrt{w}+1}\&DifferentialD;w=6\left(\genfrac{}{}{0ex}{}{\frac{u^5}{5}}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\frac{u^5}{5}}}{w=a..b}\right)-6\left(\genfrac{}{}{0ex}{}{\frac{u^2}{2}}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\frac{u^2}{2}}}{w=a..b}\right)+\genfrac{}{}{0ex}{}{\ln \left(u^2-u+1\right)}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\ln \left(u^2-u+1\right)}}{w=a..b}+12\left(\genfrac{}{}{0ex}{}{\frac{\sqrt{3}\arctan \left(\frac{2\left(u-\frac{1}{2}\right)\sqrt{3}}{3}\right)}{6}}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\frac{\sqrt{3}\arctan \left(\frac{2\left(u-\frac{1}{2}\right)\sqrt{3}}{3}\right)}{6}}}{w=a..b}\right)-2\left(\genfrac{}{}{0ex}{}{\ln \left(u+1\right)}{\phantom{w=a..b}}|\genfrac{}{}{0ex}{}{\phantom{\ln \left(u+1\right)}}{w=a..b}\right)$

$\mathrm{Rule}\left[\mathrm{revert}\right]\left(\right)$

Reverting substitution using u = w^(1/6)

${\int }_a^b\frac{w^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{\sqrt{w}+1}\&DifferentialD;w=\frac{6b^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}{5}-\frac{6a^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}{5}-3b^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+3a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+\ln \left(b^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-b^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}+1\right)-\ln \left(a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}+1\right)+2\sqrt{3}\arctan \left(\frac{\sqrt{3}\left(2b^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}-1\right)}{3}\right)-2\sqrt{3}\arctan \left(\frac{\sqrt{3}\left(2a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}-1\right)}{3}\right)-2\ln \left(b^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}+1\right)+2\ln \left(a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}+1\right)$

$\&int;\sqrt{1-x^2}\&DifferentialD;x$

$\int \sqrt{-x^2+1}\&DifferentialD;x$

$\mathrm{Rule}\left[\mathrm{change}\&comma;x\&equals;\tan \left(u\right)\right]\left(\right)$

Creating problem #9

Applying substitution x = tan(u), u = arctan(x) with dx = (1+tan(u)^2)*du, du = 1/(x^2+1)*dx

$\int \sqrt{-x^2+1}\&DifferentialD;x=\int \sqrt{-{\tan \left(u\right)}^2+1}{\tan \left(u\right)}^2\&DifferentialD;u+\int \sqrt{-{\tan \left(u\right)}^2+1}\&DifferentialD;u$

$\mathrm{Undo}\left(\right)$

$\int \sqrt{-x^2+1}\&DifferentialD;x$

$\mathrm{Rule}\left[\mathrm{change}\&comma;x\&equals;\sin \left(u\right)\right]\left(\right)$

Applying substitution x = sin(u), u = arcsin(x) with dx = cos(u)*du, du = 1/(-x^2+1)^(1/2)*dx

$\int \sqrt{-x^2+1}\&DifferentialD;x=u-\left(\int {\sin \left(u\right)}^2\&DifferentialD;u\right)$

$\mathrm{DiffTutor}\left(\right)$

$\frac{\&DifferentialD;}{\&DifferentialD;x}\left(x\sin \left(x\right)\right)$

$\frac{\&DifferentialD;}{\&DifferentialD;x}\left(x\sin \left(x\right)\right)$

$\mathrm{IntTutor}\left(\right)$

$\int {\sin \left(x\right)}^2\&DifferentialD;x$

$\int {\sin \left(x\right)}^2\&DifferentialD;x$

$\mathrm{LimitTutor}\left(\right)$

$\underset{x\to 0}{lim}x\cos \left(x\right)\ln \left(x\right)$

$\underset{x\to 0}{lim}x\cos \left(x\right)\ln \left(x\right)$