What is the integral of x/((x+2)^{1/4)} ?

The integral of x/((x+2)^{1/4)} is 4/7 (x+2)^{7/4}-8/3 (x+2)^{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 x/((x+2)^{1/4)}

\int \frac{x}{( x + 2 ) ^{\frac{1}{4}}} d x

\frac{4}{7} (x + 2)^{\frac{7}{4}} - \frac{8}{3} (x + 2)^{\frac{3}{4}} + C

Solution steps

\int \frac{x}{( x + 2 ) ^{\frac{1}{4}}} d x

Apply u-substitution

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

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

Expand

\frac{u - 2}{u ^{\frac{1}{4}}} : u^{\frac{3}{4}} - \frac{2}{u ^{\frac{1}{4}}}

= \int u^{\frac{3}{4}} - \frac{2}{u ^{\frac{1}{4}}} 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

= \int u^{\frac{3}{4}} d u - \int \frac{2}{u ^{\frac{1}{4}}} d u

\int u^{\frac{3}{4}} d u = \frac{4}{7} u^{\frac{7}{4}}

\int \frac{2}{u ^{\frac{1}{4}}} d u = \frac{8}{3} u^{\frac{3}{4}}

= \frac{4}{7} u^{\frac{7}{4}} - \frac{8}{3} u^{\frac{3}{4}}

Substitute back

u = x + 2

= \frac{4}{7} (x + 2)^{\frac{7}{4}} - \frac{8}{3} (x + 2)^{\frac{3}{4}}

Add a constant to the solution

= \frac{4}{7} (x + 2)^{\frac{7}{4}} - \frac{8}{3} (x + 2)^{\frac{3}{4}} + C

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

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

\int \frac{9 x ^{4}}{4 x ^{5} + 3} d x

\int 28 x^{7} d x

\int 23 e^{u} d u

What is the integral of x/((x+2)^{1/4)} ?
The integral of x/((x+2)^{1/4)} is 4/7 (x+2)^{7/4}-8/3 (x+2)^{3/4}+C