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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Apply u-substitution

= \int_{x + 1}^{x + 2} \frac{1}{u ^{2}} d u

Apply the Power Rule

= [- \frac{1}{u}]_{x + 1}^{x + 2}

Compute the boundaries:

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

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

Simplify

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

\int_{- π}^{0} sin (x) cos (n x) d x

\int_{0}^{\frac{π}{2}} sin^{2} (z) d z

\int_{0}^{7} 14 x - x^{2} d x

\int_{0}^{6} \frac{x}{x ^{2} + 1} d x

\int_{0}^{1} 2 x^{3} 1 - x^{2} d x

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