What is the integral of (3x)/(sqrt(4x^2+5)) ?

The integral of (3x)/(sqrt(4x^2+5)) is 3/4 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 (3x)/(sqrt(4x^2+5))

\int \frac{3 x}{4 x ^{2} + 5} d x

\frac{3}{4} 4 x^{2} + 5 + C

Solution steps

\int \frac{3 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

= 3 \cdot \int \frac{x}{4 x ^{2} + 5} d x

Apply u-substitution

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

= 3 \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

= 3 \cdot \frac{1}{8} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= 3 \cdot \frac{1}{8} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 4 x^{2} + 5

= 3 \cdot \frac{1}{8} \cdot 2 (4 x^{2} + 5)^{\frac{1}{2}}

Simplify

3 \cdot \frac{1}{8} \cdot 2 (4 x^{2} + 5)^{\frac{1}{2}} : \frac{3}{4} 4 x^{2} + 5

= \frac{3}{4} 4 x^{2} + 5

Add a constant to the solution

= \frac{3}{4} 4 x^{2} + 5 + C

(8 x + 9 y) d x + (9 x + 6 y) d y = 0

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

(x + 1) y^{^{'}} + (x + 2) y = 2 x e^{- a x}

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

\frac{d}{d x} (x (x^{4} + 8)^{3})

What is the integral of (3x)/(sqrt(4x^2+5)) ?
The integral of (3x)/(sqrt(4x^2+5)) is 3/4 sqrt(4x^2+5)+C