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

The integral of 1/((x+1)^3*sqrt(x^2+2x)) is 1/2 (arcsec(x+1)+1/2 sin(2arcsec(x+1)))+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+1)^3*sqrt(x^2+2x))

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

\frac{1}{2} (arcsec (x + 1) + \frac{1}{2} sin (2 arcsec (x + 1))) + C

Solution steps

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

Complete the square

x^{2} + 2 x : (x + 1)^{2} - 1

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

Apply u-substitution

: \int \frac{1}{u ^{3} u ^{2} - 1} d u

= \int \frac{1}{u ^{3} u ^{2} - 1} d u

Apply Trigonometric Substitution

: \int cos^{2} (v) d v

= \int cos^{2} (v) d v

Rewrite using trig identities

= \int \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 + cos (2 v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

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

Add a constant to the solution

= \frac{1}{2} (arcsec (x + 1) + \frac{1}{2} sin (2 arcsec (x + 1))) + C

\int \frac{x ^{2} - 4 x + 4}{x ^{2} - 4 x + 8} d x

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

\int \frac{5}{p ^{3}} - \frac{p ^{3}}{5} + p d p

\int \frac{9}{36 - 81 x ^{2}} d x

\int (t^{5} + e^{3 t}) d t

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