Question

$\int \frac{{(x+\sqrt{1+x^2})}^{15}}{\sqrt{1+x^2}}dx$ is equal to

• A. $\frac{{(x+\sqrt{1+x^2})}^{16}}{10}+C$
• B. $\frac{1}{15\left(\sqrt{1+x^2+x}\right)}+C$
• C. $\frac{15}{(\sqrt{1+x^2}-x)}+C$
• D. $\frac{{(x+\sqrt{1+x^2})}^{15}}{15}+C$

Answer 1

The correct option is D $\frac{{(x+\sqrt{1+x^2})}^{15}}{15}+C$

$\int \frac{{(x+\sqrt{1+x^2})}^{15}}{\sqrt{1+x^2}}dx$
$=\int \frac{{(x+\sqrt{1+x^2})(x+\sqrt{1+x^2})}^{14}}{\sqrt{1+x^2}}dx$
Substitute $(x+\sqrt{1+x^2})=t$
$(1+\frac{2x}{2\sqrt{1+x^2}})dx=dt$
$\frac{(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx=dt$
So, $I=\int t^{14}dt$
$=\frac{t^{15}}{15}+C$
$I=\frac{{(x+\sqrt{1+x^2})}^{15}}{15}+C$