What is the integral of 1/(xsqrt(6+4x)) ?

The integral of 1/(xsqrt(6+4x)) is -1/(sqrt(6))(ln|sqrt((3+2x)/3)+1|-ln|sqrt((3+2x)/3)-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(6+4x))

\int \frac{1}{x 6 + 4 x} d x

- \frac{1}{6} (ln \frac{3 + 2 x}{3} + 1 - ln \frac{3 + 2 x}{3} - 1) + C

Solution steps

\int \frac{1}{x 6 + 4 x} d x

Apply u-substitution

: \int \frac{2}{u ^{2} - 6} d u

= \int \frac{2}{u ^{2} - 6} 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} - 6} d u

Apply Integral Substitution

= 2 \cdot \int \frac{1}{6 ( 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}{6} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

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

= 2 \cdot \frac{1}{6} \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}{6} (- \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}{6} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 2 \cdot \frac{1}{6} - \frac{ln \frac{6 + 4 x}{6} + 1}{2} - \frac{ln \frac{6 + 4 x}{6} - 1}{2}

Simplify

2 \cdot \frac{1}{6} - \frac{ln \frac{6 + 4 x}{6} + 1}{2} - \frac{ln \frac{6 + 4 x}{6} - 1}{2} : - \frac{1}{6} (ln \frac{3 + 2 x}{3} + 1 - ln \frac{3 + 2 x}{3} - 1)

= - \frac{1}{6} (ln \frac{3 + 2 x}{3} + 1 - ln \frac{3 + 2 x}{3} - 1)

Add a constant to the solution

= - \frac{1}{6} (ln \frac{3 + 2 x}{3} + 1 - ln \frac{3 + 2 x}{3} - 1) + C

f (x) = x^{2} (2 x + 9)^{3}

\int_{0}^{π} cos (4 x) d x

\frac{d h}{d t} + 8 \cdot h = 4

\frac{d y}{d x} = 6 - y

\int (4 e^{x}) d x

What is the integral of 1/(xsqrt(6+4x)) ?
The integral of 1/(xsqrt(6+4x)) is -1/(sqrt(6))(ln|sqrt((3+2x)/3)+1|-ln|sqrt((3+2x)/3)-1|)+C