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

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

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

- \frac{1}{y + 2} + C

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = y + 2

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

Add a constant to the solution

= - \frac{1}{y + 2} + C

\frac{d y}{d x} + 3 x^{2} y = x

x \to 1 lim (x^{4} - x^{2} + 2)

y^{^{'}} + 0.4 \cdot y = 0.7 \cdot (299 \cdot e^{- 0.7 \cdot x})

x \to 1 + lim (2^{\frac{1}{x - 1}})

x (2 y - 3) d x + (x^{2} + 1) d y = 0