What is the integral of (16x)/(x^2+1) ?

The integral of (16x)/(x^2+1) is 8ln|x^2+1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (16x)/(x^2+1)

\int \frac{16 x}{x ^{2} + 1} d x

8 ln x^{2} + 1 + C

Solution steps

\int \frac{16 x}{x ^{2} + 1} d x

Take the constant out:

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

= 16 \cdot \int \frac{x}{x ^{2} + 1} d x

Apply u-substitution

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

= 16 \cdot \int \frac{1}{2 u} d u

Take the constant out:

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

= 16 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 16 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 1

= 16 \cdot \frac{1}{2} ln x^{2} + 1

Simplify

16 \cdot \frac{1}{2} ln x^{2} + 1 : 8 ln x^{2} + 1

= 8 ln x^{2} + 1

Add a constant to the solution

= 8 ln x^{2} + 1 + C

\int \frac{ln ( 2 x )}{x} d x

\int \frac{y ^{2}}{( 16 y ^{2} + 9 ) ^{\frac{3}{2}}} d y

\int t \frac{1}{4} d t

\int \frac{6 x + 1}{4 x ^{2} + 4 x + 10} d x

\int \frac{3 x ^{2} - 1}{( x ^{3} - x ) ^{2}} d x

What is the integral of (16x)/(x^2+1) ?
The integral of (16x)/(x^2+1) is 8ln|x^2+1|+C