What is the integral of 1/(xsqrt(x-2)) ?

The integral of 1/(xsqrt(x-2)) is sqrt(2)arctan(sqrt(1/2 (x-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/(xsqrt(x-2))

\int \frac{1}{x x - 2} d x

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

Solution steps

\int \frac{1}{x x - 2} d x

Apply u-substitution

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

= 2 \cdot \frac{1}{2} arctan (v)

Substitute back

= 2 \cdot \frac{1}{2} arctan (\frac{x - 2}{2})

Simplify

2 \cdot \frac{1}{2} arctan (\frac{x - 2}{2}) : 2 arctan (\frac{1}{2} (x - 2))

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

Add a constant to the solution

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

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{2 x}

\int \frac{1}{x + 1} d x

x e^{x^{2}}

\int_{1}^{x} sec (t) d t

\frac{d}{d x} (\frac{10}{27 x ^{\frac{8}{3}}})

What is the integral of 1/(xsqrt(x-2)) ?
The integral of 1/(xsqrt(x-2)) is sqrt(2)arctan(sqrt(1/2 (x-2)))+C