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Practicing Success
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$\int\frac{(x+\sqrt{1+x^2})^{2009}}{\sqrt{1+x^2}}dx$ is equal to : |
$\frac{2(x+\sqrt{1+x^2})^{2009}}{2009}+C$ $\frac{(x+\sqrt{1+x^2})^{2009}}{2009}+C$ $\frac{(\sqrt{1+x^2})^{2009}}{2009}+C$ None of these |
$\frac{(x+\sqrt{1+x^2})^{2009}}{2009}+C$ |
Put $x+\sqrt{1+x^2}=t⇒1+\frac{2x}{2\sqrt{1+x^2}}dx=dt⇒\frac{(\sqrt{1+x^2}+x)}{\sqrt{1+x^2}}dx=dt$ $∴I=\int\frac{t^{2009}}{t}dt=\frac{t^{2009}}{2009}+C=\frac{(x+\sqrt{1+x^2})^{2009}}{2009}+C$ |
