What is the integral of 1/((x^2+4x+8)^{3/2)} ?

The integral of 1/((x^2+4x+8)^{3/2)} is (x+2)/(4(x^2+4x+8)^{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+4x+8)^{3/2)}

\int \frac{1}{( x ^{2} + 4 x + 8 ) ^{\frac{3}{2}}} d x

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

Solution steps

\int \frac{1}{( x ^{2} + 4 x + 8 ) ^{\frac{3}{2}}} d x

Complete the square

x^{2} + 4 x + 8 : (x + 2)^{2} + 4

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

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

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

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 ( v )} d v

Rewrite using trig identities

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int x cos (x) y d y

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

\int \frac{ln ( 7 x )}{2 x} d x

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

\int (x - 1) d x

What is the integral of 1/((x^2+4x+8)^{3/2)} ?
The integral of 1/((x^2+4x+8)^{3/2)} is (x+2)/(4(x^2+4x+8)^{1/2)}+C