What is the integral of 6/(xsqrt(x^2+1)) ?

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

\int \frac{6}{x x ^{2} + 1} d x

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

Solution steps

\int \frac{6}{x x ^{2} + 1} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{x x ^{2} + 1} d x

Apply u-substitution

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

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

Factor

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

= 6 \cdot \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

= 6 (- \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}

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

Substitute back

u = x^{2} + 1

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

Simplify

6 (- (\frac{ln x ^{2} + 1 + 1}{2} - \frac{ln x ^{2} + 1 - 1}{2})) : - 6 (\frac{1}{2} ln x^{2} + 1 + 1 - \frac{1}{2} ln x^{2} + 1 - 1)

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

Add a constant to the solution

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

\int (x + 1) e^{9 x^{2} + 18 x} d x

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 13 y = 8 sin (3 x), y (0) = 1, y^{^{'}} (0) = 2

x \to 1 lim ((9 x^{2}) + 30 x + 25)

4 y

\int x^{25} sin (x^{26}) d x

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