What is the integral of x\sqrt[5]{81+x^2} ?

The integral of x\sqrt[5]{81+x^2} is 5/12 (81+x^2)^{6/5}+C

Popular >

integral of x\sqrt[5]{81+x^2}

\frac{5}{12} (81 + x^{2})^{\frac{6}{5}} + C

Solution steps

Apply u-substitution

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

Apply the Power Rule

: \frac{5}{6} u^{\frac{6}{5}}

= \frac{1}{2} \cdot \frac{5}{6} u^{\frac{6}{5}}

Substitute back

u = 81 + x^{2}

= \frac{1}{2} \cdot \frac{5}{6} (81 + x^{2})^{\frac{6}{5}}

Simplify

\frac{1}{2} \cdot \frac{5}{6} (81 + x^{2})^{\frac{6}{5}} : \frac{5}{12} (81 + x^{2})^{\frac{6}{5}}

= \frac{5}{12} (81 + x^{2})^{\frac{6}{5}}

Add a constant to the solution

= \frac{5}{12} (81 + x^{2})^{\frac{6}{5}} + C

\int e^{- 3 z} d z

\int 18 sec^{2} (x) tan^{5} (x) d x

\int \frac{( x ^{3} - x ^{2} ) ^{2}}{x} d x

\int cos (5 x + 4 n) sin^{2} (5 x + 4 n) d x

\int \frac{5}{8} d x

What is the integral of x\sqrt[5]{81+x^2} ?
The integral of x\sqrt[5]{81+x^2} is 5/12 (81+x^2)^{6/5}+C