Integrals

$\begin{array}{l}Put\left(1+logx\right)=t\\ \Rightarrow \frac{1}{x}dx=dt\\ I=\int \frac{{(1+logx)}^2}{x}dx\\ =\int t^{2}dt\\ =\frac{t^3}{3}+C\frac{{(1+logx)}^3}{3}+C\end{array}$

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$\begin{array}{l}I=\int \frac{1}{1-tanx}dx\\ =\int \frac{1}{1-\frac{sinx}{cosx}}dx=\int \frac{1}{\frac{cosx-sinx}{cosx}}dx\\ =\int \frac{cosx}{cosx-sinx}dx\\ =\frac{1}{2}\int \frac{2cosx}{cosx-sinx}dx\\ =\frac{1}{2}\int \frac{(cosx-sinx)+(cosx+sinx)}{(cosx-sinx)}dx\\ =\frac{1}{2}{\displaystyle \int 1}·dx+\frac{1}{2}\int \frac{cosx+sinx}{cosx-sinx}dx\\ Putcosx-sinx=t\\ \Rightarrow \left(-sinx-cosx\right)dx=dt\\ I=\frac{x}{2}+\frac{1}{2}\int \frac{-dt}{t}\\ =\frac{x}{2}-\frac{1}{2}log|cosx-sinx|+C\end{array}$

$\begin{array}{l}\int \frac{1}{1+\frac{cosx}{sinx}}·dx=\int \frac{1}{\frac{sinx+cosx}{sinx}}dx\\ =\int \frac{sinx}{sin+cosx}·dx=\frac{1}{2}\int \frac{2sinx}{sinx+cosx}dx\\ =\frac{1}{2}\int (\frac{sinx+cosx)+(sinx-cosx)}{sinx+cosx}dx\\ =\frac{1}{2}{\displaystyle \int 1}·dx+1\int \frac{\left(sinx-cosx\right)}{sinx+cosx}dx\\ =\frac{1}{2}{\displaystyle \int 1}·dx+1\int \frac{\left(sinx-cosx\right)}{sinx+cosx}dx\\ =\frac{1}{2}(x)+\frac{1}{2}\int \frac{sinx-cosx}{sinx+cosx}dx\\ Putsinx+cosx=t\\ \Rightarrow \left(cosx-sinx\right)dx=dt\\ =\frac{x}{2}+\frac{1}{2}\int \frac{-dt}{t}\\ =\frac{x}{2}-\frac{1}{2}log\left|t\right|+\; c\\ =\frac{x}{2}-\frac{1}{2}log|sinx+cosx|+\; c\end{array}$

$\begin{array}{l}Put1+cosx=t\\ -sinxdx=dt\\ I=\int \frac{sinx}{{(1+cosx)}^2}dx\\ =-\int \frac{dt}{t^2}=-\int t^{-2}dt\\ =\frac{1}{t}+C\\ =\frac{1}{1+cosx}+C\end{array}$

$\begin{array}{l}Put1+cosx=t\\ \Rightarrow -sinxdx=dt\\ I=\int \frac{sinx}{1+cosx}dx={\displaystyle \int -}\frac{dt}{t}\\ =-log\left|t\right|+c\\ =-log\left|1+cosx\right|+c\end{array}$

$\begin{array}{l}Putlogsinx=t\\ \Rightarrow \frac{1}{sinx}·cosxdx=dt\\ \Rightarrow cotxdx=tx\\ I={\displaystyle \int cot}xlogsinxdx\\ ={\displaystyle \int tdt=\frac{t^2}{2}+C=\frac{{\left(logsinx\right)}^2}{2}+C}\end{array}$

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Kindly go through the solution

Kindly go through the solution