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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

Apply u-substitution

Apply the Power Rule

: \frac{4}{3} u^{\frac{3}{4}}

= \frac{4}{3} u^{\frac{3}{4}}

Substitute back

u = x + 1

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

Add a constant to the solution

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

\int 1 + cot^{2} (5 x) d x

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

\int - 9 e^{x} d x

\int (t + 1) d t

\int \frac{sec ^{2} ( ln ( t ) )}{t} d t

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