Question

Evaluate $\int \frac{{(x+\sqrt{1+x^2})}^{15}}{\sqrt{1+x^2}}dx$.

• A. $\frac{{(x+\sqrt{1+x^2})}^{14}}{14}+C$
• B. $\frac{{(x+\sqrt{1+x^2})}^{15}}{15}+C$
• C. $\frac{{(x+\sqrt{1+x^2})}^{16}}{16}+C$
• D. $\frac{{(x+\sqrt{1+x^2})}^{17}}{17}+C$

Answer 1

Answer: (B)

$\frac{{(x+\sqrt{1+x^2})}^{15}}{15}+C$

$I=\int \frac{{(x+\sqrt{1+x^2})}^{15}}{\sqrt{1+x^2}}dx$

Put $(x+\sqrt{1+x^2})=t$

$\Rightarrow (1+\frac{x}{\sqrt{1+x^2}})dx=dt$

$\Rightarrow \frac{tdx}{\sqrt{1+x^2}}=dt$
$\therefore I=\int t^{15}.\frac{dt}{t}$

$=\int t^{14}dt$

$=\frac{t^{15}}{15}+C$

$=\frac{{(x+\sqrt{1+x^2})}^{15}}{15}+C$