What is the integral of x\sqrt[4]{9+x^2} ?

The integral of x\sqrt[4]{9+x^2} is 1/2 x^2\sqrt[4]{9+x^2}-1/10 (x^2-36)\sqrt[4]{x^2+9}+C

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integral of x\sqrt[4]{9+x^2}

Solution steps

Apply Integration By Parts

\int \frac{x ^{3}}{4 ( x ^{2} + 9 ) ^{\frac{3}{4}}} d x = \frac{1}{10} (x^{2} - 36) (x^{2} + 9)^{\frac{1}{4}}

Apply radical rule:

Add a constant to the solution

\int 32 x^{3} - 18 x^{2} + 8 x d x

\int \frac{x ^{2}}{x x ^{2} - 9} d x

\int \frac{x ^{2} - 2 x + 11}{x ^{2} + 3 x - 4} d x

\int (sec (x))^{2} tan (x) d x

\int e^{- 0.4 t} d t

What is the integral of x\sqrt[4]{9+x^2} ?
The integral of x\sqrt[4]{9+x^2} is 1/2 x^2\sqrt[4]{9+x^2}-1/10 (x^2-36)\sqrt[4]{x^2+9}+C