What is the integral of ((x+1))/(\sqrt[3]{x^2+2x+1)} ?

The integral of ((x+1))/(\sqrt[3]{x^2+2x+1)} is 3/4 (x^2+2x+1)^{2/3}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((x+1))/(\sqrt[3]{x^2+2x+1)}

\frac{3}{4} (x^{2} + 2 x + 1)^{\frac{2}{3}} + C

Solution steps

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

: \frac{3}{2} u^{\frac{2}{3}}

= \frac{1}{2} \cdot \frac{3}{2} u^{\frac{2}{3}}

Substitute back

u = x^{2} + 2 x + 1

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

Simplify

\frac{1}{2} \cdot \frac{3}{2} (x^{2} + 2 x + 1)^{\frac{2}{3}} : \frac{3}{4} (x^{2} + 2 x + 1)^{\frac{2}{3}}

= \frac{3}{4} (x^{2} + 2 x + 1)^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{4} (x^{2} + 2 x + 1)^{\frac{2}{3}} + C

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

\int \frac{x ^{2}}{9 x ^{2} + 4} d x

\int (2 x y^{2} - 3) d x

\int 64 q q^{2} + 4 d q

\int \frac{1}{4 x ^{2} + 4 x - 3} d x

What is the integral of ((x+1))/(\sqrt[3]{x^2+2x+1)} ?
The integral of ((x+1))/(\sqrt[3]{x^2+2x+1)} is 3/4 (x^2+2x+1)^{2/3}+C