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

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

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

- \frac{x ^{2} + 2 x + 2}{x + 1} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2} u ^{2} + 1} d u

= \int \frac{1}{u ^{2} u ^{2} + 1} d u

Apply Trigonometric Substitution

: \int cot^{2} (v) sec (v) d v

= \int cot^{2} (v) sec (v) d v

Simplify

cot^{2} (v) sec (v) : cot (v) csc (v)

= \int cot (v) csc (v) d v

Apply u-substitution

: \int - 1 d w

= \int - 1 d w

Integral of a constant:

\int a d x = a x

= (- 1) w

Substitute back

= (- 1) csc (arctan (x + 1))

Simplify

(- 1) csc (arctan (x + 1)) : - \frac{x ^{2} + 2 x + 2}{x + 1}

= - \frac{x ^{2} + 2 x + 2}{x + 1}

Add a constant to the solution

= - \frac{x ^{2} + 2 x + 2}{x + 1} + C

\int 9 x (4 + 3 x)^{5} d x

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

\int 0.1682 x^{2} + 0.0987 x + 0.4774 d x

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

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

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