What is the integral of 1/(xsqrt(ax+b)) ?

The integral of 1/(xsqrt(ax+b)) is -1/(sqrt(b))(ln|sqrt((ax+b)/b)+1|-ln|sqrt((ax+b)/b)-1|)+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(ax+b))

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

- \frac{1}{b} (ln \frac{a x + b}{b} + 1 - ln \frac{a x + b}{b} - 1) + C

Solution steps

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

Apply u-substitution

: \int \frac{2}{u ^{2} - b} d u

= \int \frac{2}{u ^{2} - b} 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} - b} d u

Apply Integral Substitution

= 2 \cdot \int \frac{1}{b ( 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}{b} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

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

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 2 \cdot \frac{1}{b} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 2 \cdot \frac{1}{b} - \frac{ln \frac{a x + b}{b} + 1}{2} - \frac{ln \frac{a x + b}{b} - 1}{2}

Simplify

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

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

Add a constant to the solution

= - \frac{1}{b} (ln \frac{a x + b}{b} + 1 - ln \frac{a x + b}{b} - 1) + C

f (x) = \frac{20 x}{x ^{2} - 5}

\frac{2 x ^{2} + 5 x}{x}

\frac{2}{s + 2}

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

R (x) = [x^{2} (x^{2} + 6 x)]^{4}

What is the integral of 1/(xsqrt(ax+b)) ?
The integral of 1/(xsqrt(ax+b)) is -1/(sqrt(b))(ln|sqrt((ax+b)/b)+1|-ln|sqrt((ax+b)/b)-1|)+C