What is the integral of 1/((x^2+2^2)^{3/2)} ?

The integral of 1/((x^2+2^2)^{3/2)} is x/(4(4+x^2)^{1/2)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((x^2+2^2)^{3/2)}

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

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

Solution steps

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

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

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

Simplify

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

= \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{1}{2}}}

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

Add a constant to the solution

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

\int (x + x^{3}) e^{x} d x

\int \frac{1}{2 x ^{2} - 7 x + 4} d x

\int (6 - 6 csc (x) cot (x) - 6^{x}) d x

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

\int (\frac{36 x ^{3} - 109 x ^{2} + 25}{3 x - 5}) d x

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