Question

If $\int (x+\sqrt{1+x^2}{)}^{n}dx=\frac{1}{a(n+1)}{\{x+\sqrt{1+x^2}\}}^{n+1}+\frac{1}{b(n-1)}{\{x+\sqrt{1+x^2}\}}^{n-1}+C;(n>1)$ then $a+b=$

Answer 1

Given :
$\int {(x+\sqrt{1+x^2})}^{n}dx=\frac{1}{a(n+1)}{\{x+\sqrt{1+x^2}\}}^{n+1}+\frac{1}{b(n-1)}{\{x+\sqrt{1+x^2}\}}^{n-1}+C\cdots \left(i\right)$
Let $I=\int (x+\sqrt{1+x^2}{)}^{n}dx$
Now substituting,
$x+\sqrt{1+x^2}=z\cdots \left(ii\right)$
$\because (\sqrt{1+x^2}+x)(\sqrt{1+x^2}-x)=1$
$\Rightarrow \sqrt{1+x^2}-x=\frac{1}{z}\cdots \left(iii\right)$
Solving $\left(ii\right)$ and $\left(iii\right)$, we get
$x=\frac{1}{2}(z-\frac{1}{z})$
$\Rightarrow dx=\frac{1}{2}(1+\frac{1}{z^2})dz$
The integral becomes:
$I=\int z^n\cdot \frac{1}{2}(1+\frac{1}{z^2})dz$
$\Rightarrow I=\frac{1}{2}\int [z^n+z^{n-2}]dz$
$\Rightarrow I=\frac{1}{2}[\frac{z^{n+1}}{n+1}+\frac{z^{n-1}}{n-1}]+C$
$\Rightarrow I=\frac{1}{2(n+1)}{\{x+\sqrt{1+x^2}\}}^{n+1}+\frac{1}{2(n-1)}{\{x+\sqrt{1+x^2}\}}^{n-1}+C$
Comparing with equation $\left(i\right),$ We get
$a=2$ and $b=2$
$\therefore a+b=4$