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

The integral of (4xln(sqrt(1+x^2)))/(1+x^2) is 1/2 ln^2(1+x^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4xln(sqrt(1+x^2)))/(1+x^2)

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

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

Solution steps

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

Take the constant out:

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

= 4 \cdot \int \frac{x ln ( 1 + x ^{2} )}{1 + x ^{2}} d x

Apply u-substitution

: \int \frac{ln ( 1 + u )}{4 ( 1 + u )} d u

= 4 \cdot \int \frac{ln ( 1 + u )}{4 ( 1 + u )} d u

Take the constant out:

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

= 4 \cdot \frac{1}{4} \cdot \int \frac{ln ( 1 + u )}{1 + u} d u

\frac{ln ( 1 + u )}{1 + u} = ln (1 + u) \frac{1}{1 + u}

= 4 \cdot \frac{1}{4} \cdot \int ln (1 + u) \frac{1}{1 + u} d u

Apply u-substitution

: \int v d v

= 4 \cdot \frac{1}{4} \cdot \int v d v

Apply the Power Rule

: \frac{v ^{2}}{2}

= 4 \cdot \frac{1}{4} \cdot \frac{v ^{2}}{2}

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

y d x + x^{2} + 1 d y = 0

\int x x - 18 d x

\frac{d}{d x} (4 e^{- 2 x^{2}})

(- 2.2) (2.3)

\frac{cos ( x )}{x ^{3}}

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