What is the integral of \sqrt[5]{t^4-t^2}(10t^3-5t) ?

The integral of \sqrt[5]{t^4-t^2}(10t^3-5t) is 25/12 (-t^2+t^4)^{6/5}+C

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integral of \sqrt[5]{t^4-t^2}(10t^3-5t)

\frac{25}{12} (- t^{2} + t^{4})^{\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{5}{2} \cdot \frac{5}{6} u^{\frac{6}{5}}

Substitute back

u = - t^{2} + t^{4}

= \frac{5}{2} \cdot \frac{5}{6} (- t^{2} + t^{4})^{\frac{6}{5}}

Simplify

\frac{5}{2} \cdot \frac{5}{6} (- t^{2} + t^{4})^{\frac{6}{5}} : \frac{25}{12} (- t^{2} + t^{4})^{\frac{6}{5}}

= \frac{25}{12} (- t^{2} + t^{4})^{\frac{6}{5}}

Add a constant to the solution

= \frac{25}{12} (- t^{2} + t^{4})^{\frac{6}{5}} + C

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What is the integral of \sqrt[5]{t^4-t^2}(10t^3-5t) ?
The integral of \sqrt[5]{t^4-t^2}(10t^3-5t) is 25/12 (-t^2+t^4)^{6/5}+C