What is the integral of ((y^5)/b+a)^{7/8}y^4 ?

The integral of ((y^5)/b+a)^{7/8}y^4 is (8(y^5+ba)^{15/8})/(75b^{7/8)}+C

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integral of ((y^5)/b+a)^{7/8}y^4

\int (\frac{y ^{5}}{b} + a)^{\frac{7}{8}} y^{4} d y

\frac{8 ( y ^{5} + ba ) ^{\frac{15}{8}}}{75 b ^{\frac{7}{8}}} + C

Solution steps

\int (\frac{y ^{5}}{b} + a)^{\frac{7}{8}} y^{4} d y

Apply u-substitution

= \int \frac{b u ^{\frac{7}{8}}}{5} d u

Take the constant out:

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

= \frac{b}{5} \cdot \int u^{\frac{7}{8}} d u

Apply the Power Rule

= \frac{b}{5} \cdot \frac{8}{15} u^{\frac{15}{8}}

Substitute back

u = \frac{y ^{5}}{b} + a

= \frac{b}{5} \cdot \frac{8}{15} (\frac{y ^{5}}{b} + a)^{\frac{15}{8}}

Simplify

\frac{b}{5} \cdot \frac{8}{15} (\frac{y ^{5}}{b} + a)^{\frac{15}{8}} : \frac{8 ( y ^{5} + ba ) ^{\frac{15}{8}}}{75 b ^{\frac{7}{8}}}

= \frac{8 ( y ^{5} + ba ) ^{\frac{15}{8}}}{75 b ^{\frac{7}{8}}}

Add a constant to the solution

= \frac{8 ( y ^{5} + ba ) ^{\frac{15}{8}}}{75 b ^{\frac{7}{8}}} + C

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What is the integral of ((y^5)/b+a)^{7/8}y^4 ?
The integral of ((y^5)/b+a)^{7/8}y^4 is (8(y^5+ba)^{15/8})/(75b^{7/8)}+C