What is the integral of 1/(sqrt(a+bx)) ?

The integral of 1/(sqrt(a+bx)) is x/(sqrt(a+bx))+(sqrt(a+bx))/b+a/(bsqrt(a+bx))+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(a+bx))

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

\frac{x}{a + b x} + \frac{a + b x}{b} + \frac{a}{b a + b x} + C

Solution steps

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

Apply Integration By Parts

: \frac{x}{a + b x} - \int - \frac{b x}{2 ( a + b x ) ^{\frac{3}{2}}} d x

= \frac{x}{a + b x} - \int - \frac{b x}{2 ( a + b x ) ^{\frac{3}{2}}} d x

\int - \frac{b x}{2 ( a + b x ) ^{\frac{3}{2}}} d x = - \frac{( a + b x ) ^{\frac{1}{2}}}{b} - \frac{a}{b a + b x}

= \frac{x}{a + b x} - (- \frac{( a + b x ) ^{\frac{1}{2}}}{b} - \frac{a}{b a + b x})

Simplify

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

= \frac{x}{a + b x} + \frac{a + b x}{b} + \frac{a}{b a + b x}

Add a constant to the solution

= \frac{x}{a + b x} + \frac{a + b x}{b} + \frac{a}{b a + b x} + C

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

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

\int \frac{( 3 x ^{2} - 2 )}{( 2 x ^{3} - 4 x )} d x

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

\int (10 x^{3} - 2 sec^{2} (x)) d x

What is the integral of 1/(sqrt(a+bx)) ?
The integral of 1/(sqrt(a+bx)) is x/(sqrt(a+bx))+(sqrt(a+bx))/b+a/(bsqrt(a+bx))+C