int frac{x^{9/2}}{sqrt{1+x^{11}}} dx is equal | vedclass

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Question

(A) Let $I = \int \frac{x^{9/2}}{\sqrt{1+x^{11}}} dx$. Substitute $t = x^{11/2}$. Then, $dt = \frac{11}{2} x^{9/2} dx$, which implies $x^{9/2} dx = \f

Vedclass · Accepted Answer

$\frac{2}{11} \log \left(x^{11/2}+\sqrt{1+x^{11}}\right)+c$

Vedclass · Answer

(A) Let $I = \int \frac{x^{9/2}}{\sqrt{1+x^{11}}} dx$. Substitute $t = x^{11/2}$. Then, $dt = \frac{11}{2} x^{9/2} dx$, which implies $x^{9/2} dx = \frac{2}{11} dt$. Substituting these into the integral: $I = \int \frac{\frac{2}{11} dt}{\sqrt{1+t^2}} = \frac{2}{11} \int \frac{dt}{\sqrt{1+t^2}}$. Using the standard integral formula $\int \frac{dx}{\sqrt{1+x^2}} = \log(x + \sqrt{1+x^2}) + c$: $I = \frac{2}{11} \log(t + \sqrt{1+t^2}) + c$. Substituting back $t = x^{11/2}$: $I = \frac{2}{11} \log(x^{11/2} + \sqrt{1+x^{11}}) + c$.

$\int \frac{x^{9/2}}{\sqrt{1+x^{11}}} dx$ is equal to -

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