What is the integral of (5x-y^2)^{15} ?

The integral of (5x-y^2)^{15} is 1/80 (-y^2+5x)^{16}+C

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integral of (5x-y^2)^{15}

\int (5 x - y^{2})^{15} d x

\frac{1}{80} (- y^{2} + 5 x)^{16} + C

Solution steps

\int (5 x - y^{2})^{15} d x

Apply u-substitution

= \int \frac{u ^{15}}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int u^{15} d u

Apply the Power Rule

= \frac{1}{5} \cdot \frac{u ^{16}}{16}

Substitute back

u = - y^{2} + 5 x

= \frac{1}{5} \cdot \frac{( - y ^{2} + 5 x ) ^{16}}{16}

Simplify

\frac{1}{5} \cdot \frac{( - y ^{2} + 5 x ) ^{16}}{16} : \frac{1}{80} (- y^{2} + 5 x)^{16}

= \frac{1}{80} (- y^{2} + 5 x)^{16}

Add a constant to the solution

= \frac{1}{80} (- y^{2} + 5 x)^{16} + C

What is the integral of (5x-y^2)^{15} ?
The integral of (5x-y^2)^{15} is 1/80 (-y^2+5x)^{16}+C