What is the integral of y^6(y^2+4y-5) ?

The integral of y^6(y^2+4y-5) is (y^9)/9+(y^8)/2-(5y^7)/7+C

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integral of y^6(y^2+4y-5)

\int y^{6} (y^{2} + 4 y - 5) d y

\frac{y ^{9}}{9} + \frac{y ^{8}}{2} - \frac{5 y ^{7}}{7} + C

Solution steps

\int y^{6} (y^{2} + 4 y - 5) d y

Expand

y^{6} (y^{2} + 4 y - 5) : y^{8} + 4 y^{7} - 5 y^{6}

= \int y^{8} + 4 y^{7} - 5 y^{6} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int y^{8} d y + \int 4 y^{7} d y - \int 5 y^{6} d y

\int y^{8} d y = \frac{y ^{9}}{9}

\int 4 y^{7} d y = \frac{y ^{8}}{2}

\int 5 y^{6} d y = \frac{5 y ^{7}}{7}

= \frac{y ^{9}}{9} + \frac{y ^{8}}{2} - \frac{5 y ^{7}}{7}

Add a constant to the solution

= \frac{y ^{9}}{9} + \frac{y ^{8}}{2} - \frac{5 y ^{7}}{7} + C

What is the integral of y^6(y^2+4y-5) ?
The integral of y^6(y^2+4y-5) is (y^9)/9+(y^8)/2-(5y^7)/7+C