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

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

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

2 ln 1 + x^{2} + C

Solution steps

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

Apply u-substitution

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

= 4 \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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 + x^{2}

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

Simplify

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

= 2 ln 1 + x^{2}

Add a constant to the solution

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

3 x y^{^{'}} - 2 y = x^{3} y^{- 2}

\frac{\partial}{\partial y} (x (e^{y} + e^{- y}))

\int \frac{- 6 x ^{2}}{( 2 - 2 x ^{3} ) ^{2}} d x

\int sin^{3} (d) x d x

f (x) = (x^{3} - 16 x)^{10}, (4, 0)

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