Question

Integrate.

$\displaystyle \int \dfrac{x^2}{1+x^5} dx$

Answer 1

GIVEN :-

$\displaystyle \int \dfrac{x^2}{1+x^5} dx$

TO FIND :-

$\begin{gathered}\begin{gathered}\begin{gathered}\\ \tt \bullet \: The \: Integrate\\ \\\end{gathered}\end{gathered} \end{gathered}$

SOLUTION :-

$\displaystyle \int \dfrac{x^2}{1+x^5} dx$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\\ : \implies{ \tt{ \int \dfrac{e^{2x}}{e^{2x}+1} - \frac{2 {e}^{x} }{ {e}^{2x} + 1} \: dx }}\\ \\ \end{gathered}\end{gathered}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\\ : \implies{ \tt{ \int \dfrac{e^{2x}}{e^{2x}+1} \: dx - \int\frac{2 {e}^{x} }{ {e}^{2x} + 1} \: dx }}\\ \\ \end{gathered}\end{gathered}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\\ : \implies{ \tt{ \frac{1}{2} \times in \bigg( {e}^{2x} + 1 \bigg) - \int\frac{2 {e}^{x} }{ {e}^{2x} + 1} \: dx }}\\ \\ \end{gathered}\end{gathered}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\\ : \implies{ \tt{ \frac{1}{2} \times in \bigg( {e}^{2x} + 1 \bigg) - 2 \: arctan \bigg( {e}^{x} \bigg) }}\\ \\ \end{gathered}\end{gathered}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\\ : \implies \large{ \boxed{ \tt{ \frac{1}{2} \times in \bigg( {e}^{2x} + 1 \bigg) - 2 \: arctan \bigg( {e}^{x} \bigg) + C}}} \\\end{gathered} \\ \\\end{gathered}\end{gathered} \\ \end{gathered}$

Answer 2

Step-by-step explanation:

Let I=∫1(x−5)2dx

=∫(x−5)−2dx

=(x−5)−2+1(−2+1)+c

= x−5^-1/−1 + C

= - 1/x−5 + C