$\int\frac{(x+\sqrt{1+x^2})^{2009}}{\sqrt{1+x^2}}dx$ is equal to :
Answer & explanation
Correct answer: option 2
Put $x+\sqrt{1+x^2}=t⇒1+\frac{2x}{2\sqrt{1+x^2}}dx=dt⇒\frac{(\sqrt{1+x^2}+x)}{\sqrt{1+x^2}}dx=dt$
$∴I=\int\frac{t^{2009}}{t}dt=\frac{t^{2009}}{2009}+C=\frac{(x+\sqrt{1+x^2})^{2009}}{2009}+C$