What is the integral of y/((y^2-4)^{2/5)} ?

The integral of y/((y^2-4)^{2/5)} is 5/6 (y^2-4)^{3/5}+C

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integral of y/((y^2-4)^{2/5)}

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

\frac{5}{6} (y^{2} - 4)^{\frac{3}{5}} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 ( u - 4 ) ^{\frac{2}{5}}} d u

= \int \frac{1}{2 ( u - 4 ) ^{\frac{2}{5}}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{( u - 4 ) ^{\frac{2}{5}}} d u

Apply u-substitution

: \int \frac{1}{v ^{\frac{2}{5}}} d v

= \frac{1}{2} \cdot \int \frac{1}{v ^{\frac{2}{5}}} d v

Apply the Power Rule

: \frac{5}{3} v^{\frac{3}{5}}

= \frac{1}{2} \cdot \frac{5}{3} v^{\frac{3}{5}}

Substitute back

= \frac{1}{2} \cdot \frac{5}{3} (y^{2} - 4)^{\frac{3}{5}}

Simplify

\frac{1}{2} \cdot \frac{5}{3} (y^{2} - 4)^{\frac{3}{5}} : \frac{5}{6} (y^{2} - 4)^{\frac{3}{5}}

= \frac{5}{6} (y^{2} - 4)^{\frac{3}{5}}

Add a constant to the solution

= \frac{5}{6} (y^{2} - 4)^{\frac{3}{5}} + C

f (x) = 12 x - x^{3}

\int - cos (n x) d x

\frac{d y}{d x} = \frac{x ^{3} + y ^{3}}{x y ^{2}}

\int_{0}^{\frac{1}{2}} 8 x - \frac{1}{2} x d x

What is the integral of y/((y^2-4)^{2/5)} ?
The integral of y/((y^2-4)^{2/5)} is 5/6 (y^2-4)^{3/5}+C