What is the integral of (\sqrt[3]{y+1})/y ?

The integral of (\sqrt[3]{y+1})/y is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (\sqrt[3]{y+1})/y

Solution steps

Apply u-substitution

: \int \frac{3 u ^{3}}{u ^{3} - 1} d u

= \int \frac{3 u ^{3}}{u ^{3} - 1} d u

Take the constant out:

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

= 3 \cdot \int \frac{u ^{3}}{u ^{3} - 1} d u

Take the partial fraction of

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

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

Apply the Sum Rule:

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

= 3 (\int 1 d u + \int \frac{1}{3 ( u - 1 )} d u + \int \frac{- u - 2}{3 ( u ^{2} + u + 1 )} d u)

\int 1 d u = u

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

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

Substitute back

Simplify

Add a constant to the solution

\int (3 cos (x) - 7 sin (x)) d x

\int \frac{1}{x ^{2} ( 1 + x ^{2} )} d x

\int \frac{3}{4 + x} d x

\int \frac{16 - ( e ^{x} ) ^{2}}{( e ^{x} ) ^{2}} d x

\int \frac{1}{( 2 + x ^{2} )} d x

What is the integral of (\sqrt[3]{y+1})/y ?
The integral of (\sqrt[3]{y+1})/y is