What is the integral of (x^{1/3})/((x^{4/3)+9)^2} ?

The integral of (x^{1/3})/((x^{4/3)+9)^2} is

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integral of (x^{1/3})/((x^{4/3)+9)^2}

\int \frac{x ^{\frac{1}{3}}}{( x ^{\frac{4}{3}} + 9 ) ^{2}} d x

Solution steps

\int \frac{x ^{\frac{1}{3}}}{( x ^{\frac{4}{3}} + 9 ) ^{2}} d x

Apply Integration By Parts

: \frac{3 x ^{\frac{4}{3}}}{4 ( x ^{\frac{4}{3}} + 9 ) ^{2}} - \int - \frac{2 x ^{\frac{5}{3}}}{( x ^{\frac{4}{3}} + 9 ) ^{3}} d x

= \frac{3 x ^{\frac{4}{3}}}{4 ( x ^{\frac{4}{3}} + 9 ) ^{2}} - \int - \frac{2 x ^{\frac{5}{3}}}{( x ^{\frac{4}{3}} + 9 ) ^{3}} d x

Simplify

Add a constant to the solution

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What is the integral of (x^{1/3})/((x^{4/3)+9)^2} ?
The integral of (x^{1/3})/((x^{4/3)+9)^2} is