Integration by Substitution

k/6 = ∫? ^ (π/6) (x²)/ (1-x²)³/² dx x = sinθ dx = cosθdθ
=> k/6 = ∫? ^ (π/6) (sin²θ)/ (1-sin²θ)³/² * cosθdθ
=> k/6 = ∫? ^ (π/6) (sin²θ)/ (cos³θ) * cosθdθ
=> k/6 = ∫? ^ (π/6) tan²θdθ = ∫? ^ (π/6) (sec²θ-1)dθ
=> k/6 = [tanθ - θ]? ^ (π/6) = (1/√3 - π/6) = (2√3-π)/6
=> k = 2√3 - π

$f\left(x\right)={\int }_1^3?\frac{\sqrt[]{x}dx}{(1+x{)}^2}={\int }_1^{\sqrt[]{3}}?\frac{t.2tdt}{{\left(1+t^2\right)}^2}\mathrm{}$ (put $\sqrt[]{x}=t$ )

$={\left(-\frac{\mathrm{t}}{1+{\mathrm{t}}^2}\right)}_1^{\sqrt[]{3}}+{\left({\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?\mathrm{t}\right)}_1^{\sqrt[]{3}}\mathrm{}$ [Applying by parts]

$\begin{array}{rr}& =-\left(\frac{\sqrt[]{3}}{4}-\frac{1}{2}\right)+\frac{\pi }{3}-\frac{\pi }{4}\\ & =\frac{1}{2}-\frac{\sqrt[]{3}}{4}+\frac{\pi }{12}\end{array}$

 $\int {\left(\frac{x}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)}^{2}dx=\int \left(\frac{x}{\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)\cdot \frac{x\mathrm{c}\mathrm{o}\mathrm{s}?xdx}{(x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x{)}^2}$

$\begin{array}{rr}& =\frac{x}{\mathrm{c}\mathrm{o}\mathrm{s}?x}\left(-\frac{1}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)+\int \left(\frac{\mathrm{c}\mathrm{o}\mathrm{s}?x+x\mathrm{s}\mathrm{i}\mathrm{n}?x}{{\mathrm{c}\mathrm{o}\mathrm{s}}^2?x}\right)\left(\frac{1}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)dx\\ & =-\frac{x\mathrm{s}\mathrm{e}\mathrm{c}?x}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}+\int {\mathrm{s}\mathrm{e}\mathrm{c}}^2?xdx\\ & =-\frac{x\mathrm{s}\mathrm{e}\mathrm{c}?x}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}+\mathrm{t}\mathrm{a}\mathrm{n}?x+C\end{array}$