What is the integral of 1/(sqrt((x^2+4)^3)) ?

The integral of 1/(sqrt((x^2+4)^3)) is x/(4sqrt(4+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 1/(sqrt((x^2+4)^3))

\int \frac{1}{( x ^{2} + 4 ) ^{3}} d x

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

Solution steps

\int \frac{1}{( x ^{2} + 4 ) ^{3}} d x

(x^{2} + 4)^{3} = (x^{2} + 4)^{\frac{3}{2}},

assuming

(x^{2} + 4) \geq 0

= \int \frac{1}{( x ^{2} + 4 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

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

= \int \frac{1}{4 sec ( u )} d u

Take the constant out:

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

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

Rewrite using trig identities

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

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{1}{4} sin (u)

Substitute back

u = arctan (\frac{1}{2} x)

= \frac{1}{4} sin (arctan (\frac{1}{2} x))

Simplify

\frac{1}{4} sin (arctan (\frac{1}{2} x)) : \frac{x}{4 4 + x ^{2}}

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

Add a constant to the solution

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

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

\int \frac{6 x + 2}{( x + 1 ) ( 2 x + 1 )} d x

\int 2 e^{x} 1 - e^{2 x} d x

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

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

What is the integral of 1/(sqrt((x^2+4)^3)) ?
The integral of 1/(sqrt((x^2+4)^3)) is x/(4sqrt(4+x^2))+C