What is the integral of 1/((y+4)^2) ?

The integral of 1/((y+4)^2) is -1/(y+4)+C

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integral of 1/((y+4)^2)

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

- \frac{1}{y + 4} + C

Solution steps

\int \frac{1}{( y + 4 ) ^{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 + 4

= - \frac{1}{y + 4}

Add a constant to the solution

= - \frac{1}{y + 4} + C

x \to \infty lim (4 x + 1)

3 - x^{2}

y = 2 x^{2} - 13 x + 3

\int x^{2} 2^{x^{3} + 1} d x

\int u^{- 2} d u