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

The integral of 1/(x^{3/2)+x^{1/2}} is 2arctan(x^{1/2})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x^{3/2)+x^{1/2}}

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

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

Solution steps

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

Multiply by the conjugate of

x^{\frac{3}{2}} + x^{\frac{1}{2}} : \frac{x ^{3} - x}{x ^{\frac{3}{2}} - x ^{\frac{1}{2}}}

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

Simplify

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

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

Apply u-substitution

: \int \frac{2}{u ^{2} + 1} d u

= \int \frac{2}{u ^{2} + 1} d u

Take the constant out:

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

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

Use the common integral:

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

= 2 arctan (u)

Substitute back

u = x^{\frac{1}{2}}

= 2 arctan (x^{\frac{1}{2}})

Add a constant to the solution

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

f (x) = \frac{x - 1}{x ^{2}}

\frac{\partial}{\partial x} (3 x^{2} y + 2 x y^{3})

n = 0 \sum \infty e^{- 1}

\int_{- 1}^{1} x e^{x} d x

\frac{d}{d x} (ln (x^{2} + 3 x))

What is the integral of 1/(x^{3/2)+x^{1/2}} ?
The integral of 1/(x^{3/2)+x^{1/2}} is 2arctan(x^{1/2})+C