Integrals

$\begin{array}{l}Put9-4x^2=t\\ -8xdx=dt\\ xdx=\frac{-1}{8}dt\\ -\frac{1}{8}{\displaystyle \int \frac{dt}{t}}=-\frac{1}{8}log\left|t\right|+c=-\frac{1}{8}log\left|9-4x^2\right|+c\end{array}$

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

$\begin{array}{l}Put,2+3x^3=t\\ 9x^{2}dx=dt\\ I=\int \frac{x^3}{{\left(2+3x^3\right)}^3}dx=\frac{1}{9}\int \frac{dt}{t^3}\\ =\frac{1}{9}\left[\frac{t^{-2}}{-2}\right]+C\\ =\frac{-1}{18}\left(\frac{1}{t^2}\right)+C=\frac{-1}{18{\left(2+3x^3\right)}^2}+C\end{array}$

$\begin{array}{l}Put,x^3-1=t\\ 3x^{2}dx=dt\\ =I\int {\left(x^3-1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{x}^{5}dx\\ =\int {\left(x^3-1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}·x^3·x^2·dx\\ =\int t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}(t+1)\frac{dt}{3}=\frac{1}{3}\int \left(t^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)dt\\ =\frac{1}{3}\left[\frac{t^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+\frac{t^{4/3}}{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right]+C\\ =\frac{1}{3}\left[\frac{3}{7}{t}^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+\frac{3}{4}{t}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right]+C\\ =\frac{1}{7}{\left(x^3-1\right)}^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+\frac{1}{4}{\left(x^3-1\right)}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+C\end{array}$

Kindly go through the solution

 

$\begin{array}{l}\frac{x-\surd x}{}=\frac{1}{\surd x\; (\surd x-1)}\\ Put,\; \surd x-1=t\\ \frac{1}{\mathrm{2\surd x}}dx=dt\\ I=\int \frac{1}{\surd x\; (\surd x-1)}dx=\int \frac{2}{t}dt\\ =2log\left|t\right|+c\\ =2log(\surd x-1)+c\end{array}$

Kindly go through the solution

$\begin{array}{l}Put,1+2x^2=t\\ dx2.2x=dt\\ dx.4x=dt\end{array}$

$\begin{array}{l}Put,x+2=t\\ dx=dt\\ \Rightarrow x=t-2\\ I={\displaystyle \int \mathrm{x\surd x+2}}dx\\ ={\displaystyle \int (}t-2)\text{\hspace{0.17em}√t}dt\\ =\int \left(t^{1+\frac{3}{2}}-2t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)dt\\ =\int \left(t^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-2t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)dt\\ =\int t^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}·dt-{\displaystyle \int 2}·t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}·dt\\ =\frac{t^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-2\frac{t^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =\frac{2}{5}{t}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{4}{3}{t}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+C\\ =\frac{2}{5}{(x+2)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{4}{3}{(x+2)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+C\end{array}$

$\begin{array}{l}Put,ax+b=t\\ adx=dt\\ dx=\frac{1}{a}dt\\ I={\displaystyle \int (}ax+b){}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}dx\\ =\int t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}·\frac{1}{a}dt=\frac{1}{a}\frac{t\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2+1$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2+1$}\right.}+C\\ =\frac{1}{a}\frac{t^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+C\\ =\frac{2}{3a}{(ax+b)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\; c\end{array}$