Question

integrate the given question by substitution method ​

Answer 1

Solution:

Given Integral:

$\displaystyle\rm\longrightarrow I=\int\dfrac{2x}{2e^{x^{2}}}\: dx$

Let us assume that:

$\rm\longrightarrow u=x^{2}$

$\rm\longrightarrow du=2x\: dx$

Therefore, our integral becomes:

$\displaystyle\rm\longrightarrow I=\int\dfrac{2x\: dx}{2e^{u}}$

$\displaystyle\rm\longrightarrow I=\int\dfrac{du}{2e^{u}}$

$\displaystyle\rm\longrightarrow I=\int\dfrac{1}{2}\dfrac{du}{e^{u}}$

$\displaystyle\rm\longrightarrow I=\dfrac{1}{2}\int e^{-u}\: du$

$\displaystyle\rm\longrightarrow I=-\dfrac{1}{2}e^{-u} + C$

Substitute back u = x². We get:

$\displaystyle\rm\longrightarrow I=\dfrac{-e^{-x^{2}}}{2} + C$

Therefore:

$\displaystyle\rm\longrightarrow \int\dfrac{2x}{2e^{x^{2}}}\: dx=\dfrac{-e^{-x^{2}}}{2} + C$

→ Which is our required answer.

Learn More:

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$