What is the integral of (4x)/(sqrt(4x^2+5)) ?

The integral of (4x)/(sqrt(4x^2+5)) is sqrt(4x^2+5)+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+5))

\int \frac{4 x}{4 x ^{2} + 5} d x

4 x^{2} + 5 + C

Solution steps

\int \frac{4 x}{4 x ^{2} + 5} 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} + 5} 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} + 5

= 4 \cdot \frac{1}{8} \cdot 2 (4 x^{2} + 5)^{\frac{1}{2}}

Simplify

4 \cdot \frac{1}{8} \cdot 2 (4 x^{2} + 5)^{\frac{1}{2}} : 4 x^{2} + 5

= 4 x^{2} + 5

Add a constant to the solution

= 4 x^{2} + 5 + C

\frac{d}{d x} (a x cos (x) + b cos (x))

\frac{d}{d x} (e^{- (x^{2} + y^{2} + z^{2})})

\frac{d y}{d x} = \frac{( x ^{3} + y ^{3} )}{3 \cdot x \cdot y ^{2}}

\frac{d y}{d x} = \frac{y ^{2}}{x ^{3}}

\int 7 x^{2} 1 - x^{3} d x

What is the integral of (4x)/(sqrt(4x^2+5)) ?
The integral of (4x)/(sqrt(4x^2+5)) is sqrt(4x^2+5)+C