What is the integral of 1/((2y+5)^2) ?

The integral of 1/((2y+5)^2) is -1/(2(2y+5))+C

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integral of 1/((2y+5)^2)

\int \frac{1}{( 2 y + 5 ) ^{2}} d y

- \frac{1}{2 ( 2 y + 5 )} + C

Solution steps

\int \frac{1}{( 2 y + 5 ) ^{2}} d y

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply the Power Rule

: - \frac{1}{u}

= \frac{1}{2} (- \frac{1}{u})

Substitute back

u = 2 y + 5

= \frac{1}{2} (- \frac{1}{2 y + 5})

Simplify

\frac{1}{2} (- \frac{1}{2 y + 5}) : - \frac{1}{2 ( 2 y + 5 )}

= - \frac{1}{2 ( 2 y + 5 )}

Add a constant to the solution

= - \frac{1}{2 ( 2 y + 5 )} + C

n = 1 \sum \infty (\frac{2}{3})^{n}

\frac{d y}{d x} = \frac{2 x + sec ^{2} ( x )}{2 y}

\int_{- 1}^{2} 1 + x^{2} d x

\frac{d}{d x} (cos (x) + x sin (x))

x y^{^{'}} - y^{2} = x y^{2}

What is the integral of 1/((2y+5)^2) ?
The integral of 1/((2y+5)^2) is -1/(2(2y+5))+C