What is the integral of 1/(xsqrt(4x^2+1)) ?

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

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

- \frac{1}{2} ln 4 x^{2} + 1 + 1 + \frac{1}{2} ln 4 x^{2} + 1 - 1 + C

Solution steps

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

Apply u-substitution

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

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

Factor

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

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

Take the constant out:

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

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

Use the common integral:

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

= - (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = 4 x^{2} + 1

= - (\frac{ln 4 x ^{2} + 1 + 1}{2} - \frac{ln 4 x ^{2} + 1 - 1}{2})

Simplify

- (\frac{ln 4 x ^{2} + 1 + 1}{2} - \frac{ln 4 x ^{2} + 1 - 1}{2}) : - \frac{1}{2} ln 4 x^{2} + 1 + 1 + \frac{1}{2} ln 4 x^{2} + 1 - 1

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

Add a constant to the solution

= - \frac{1}{2} ln 4 x^{2} + 1 + 1 + \frac{1}{2} ln 4 x^{2} + 1 - 1 + C

(cosh (x))^{^{'}}

\frac{d}{d x} (25)

x ln (x^{2} + 2)

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

x \to 0 - lim (x e^{\frac{2}{x}})

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