What is the integral of a^{12}y^4-a^4y^{12} ?

The integral of a^{12}y^4-a^4y^{12} is (a^{12}y^5)/5-(a^4y^{13})/(13)+C

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integral of a^{12}y^4-a^4y^{12}

integral

a^{12} y^{4} - a^{4} y^{12}

\frac{a ^{12} y ^{5}}{5} - \frac{a ^{4} y ^{13}}{13} + C

Solution steps

\int a^{12} y^{4} - a^{4} y^{12} 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 a^{12} y^{4} d y - \int a^{4} y^{12} d y

\int a^{12} y^{4} d y = \frac{a ^{12} y ^{5}}{5}

\int a^{4} y^{12} d y = \frac{a ^{4} y ^{13}}{13}

= \frac{a ^{12} y ^{5}}{5} - \frac{a ^{4} y ^{13}}{13}

Add a constant to the solution

= \frac{a ^{12} y ^{5}}{5} - \frac{a ^{4} y ^{13}}{13} + C

What is the integral of a^{12}y^4-a^4y^{12} ?
The integral of a^{12}y^4-a^4y^{12} is (a^{12}y^5)/5-(a^4y^{13})/(13)+C