Question

$\int \frac{[x+\sqrt{a^2+x^2}{]}^n}{\sqrt{a^2+x^2}}dx(n\ne 0)=$

• A. $\frac{1}{n+1}(x+\sqrt{a^2+x^2}{)}^{n+1}+c$
• B. $\frac{1}{n.(x+\sqrt{a^2+x^2}{)}^n}+c$
• C. $(x+\sqrt{a^2+x^2}{)}^n+c$
• D. $\frac{1}{n}(x+\sqrt{a^2+x^2}{)}^n+c$

Answer 1

The correct option is D $\frac{1}{n}(x+\sqrt{a^2+x^2}{)}^n+c$

Put $x+\sqrt{a^2+x^2}=tthen[1+\frac{2x}{2\sqrt{a^2+x^2}}]dx=dt\int \frac{[x+\sqrt{a^2+x^2}{]}^n}{\sqrt{a^2+x^2}}dx=\int t^{n-1}dt=\frac{t^n}{n}+c=\frac{1}{n}(x+\sqrt{a^2+x^2}{)}^n+c$