What is the integral from-1 to 2 of x/((x+1)^2) ?

The integral from-1 to 2 of x/((x+1)^2) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-1 to 2 of x/((x+1)^2)

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

diverges

Solution steps

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

Apply u-substitution

: \int_{0}^{3} \frac{u - 1}{u ^{2}} d u

= \int_{0}^{3} \frac{u - 1}{u ^{2}} d u

Expand

\frac{u - 1}{u ^{2}} : \frac{1}{u} - \frac{1}{u ^{2}}

= \int_{0}^{3} \frac{1}{u} - \frac{1}{u ^{2}} d u

Apply the Sum Rule:

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

= \int_{0}^{3} \frac{1}{u} d u - \int_{0}^{3} \frac{1}{u ^{2}} d u

\int_{0}^{3} \frac{1}{u} d u =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int sin (n t) d t

x^{^{'}} = 6 c_{1} - \frac{3 x}{5}

\int_{0}^{1} x (1 - x^{2}) d x

f (t) = \frac{t ^{2}}{2}

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

What is the integral from-1 to 2 of x/((x+1)^2) ?
The integral from-1 to 2 of x/((x+1)^2) is diverges