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

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

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

4 x^{2} + 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{8 u} d u

= 4 \cdot \int \frac{1}{8 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}{8} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= 4 \cdot \frac{1}{8} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 4 x^{2} + 1

= 4 \cdot \frac{1}{8} \cdot 2 (4 x^{2} + 1)^{\frac{1}{2}}

Simplify

4 \cdot \frac{1}{8} \cdot 2 (4 x^{2} + 1)^{\frac{1}{2}} : 4 x^{2} + 1

= 4 x^{2} + 1

Add a constant to the solution

= 4 x^{2} + 1 + C

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

y = e^{- x} cos (2 x)

\frac{1}{s ^{4} + s ^{2}}

\int (x^{2} + 2 x) d x

\frac{d}{d t} (t e^{4 t})

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