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

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

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

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

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

Solution steps

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

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 x + 5

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

Simplify

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

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

Add a constant to the solution

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

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

\int \frac{x ^{4} + x - 36}{x ^{2} + 6} d x

11 x d y + 11 y d x + x d x = 0

\int_{2}^{4} \frac{1}{x ^{2}} d x

t^{2} y^{^{''}} + t y^{^{'}} + 16 y = 0

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