What is the integral of 3/(2+y^3) ?

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integral of 3/(2+y^3)

\int \frac{3}{2 + y ^{3}} d y

Solution steps

\int \frac{3}{2 + y ^{3}} d y

Take the constant out:

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

= 3 \cdot \int \frac{1}{2 + y ^{3}} d y

2 + y^{3} = 2 (1 + \frac{y ^{3}}{2})

= 3 \cdot \int \frac{1}{2 ( 1 + \frac{y ^{3}}{2} )} d y

Apply exponent rule:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

Take the constant out:

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

Apply u-substitution

Take the constant out:

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

Take the partial fraction of

\frac{1}{1 + u ^{3}} : \frac{1}{3 ( u + 1 )} + \frac{- u + 2}{3 ( u ^{2} - u + 1 )}

Apply the Sum Rule:

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

\int \frac{1}{3 ( u + 1 )} d u = \frac{1}{3} ln ∣ u + 1 ∣

\int \frac{- u + 2}{3 ( u ^{2} - u + 1 )} d u = \frac{1}{6} (- ln \frac{4}{3} + \frac{4 u ^{2} - 4 u}{3} + 23 arctan (\frac{2 u - 1}{3}))

Substitute back

Simplify

Add a constant to the solution

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y^{- 2}