What is the integral of (2x+3)/((x^2+3x)^2) ?

The integral of (2x+3)/((x^2+3x)^2) is -1/(x(x+3))+C

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integral of (2x+3)/((x^2+3x)^2)

\int \frac{2 x + 3}{( x ^{2} + 3 x ) ^{2}} d x

- \frac{1}{x ( x + 3 )} + C

Solution steps

\int \frac{2 x + 3}{( x ^{2} + 3 x ) ^{2}} d x

Apply u-substitution

: \int \frac{8 u}{( u ^{2} - 9 ) ^{2}} d u

= \int \frac{8 u}{( u ^{2} - 9 ) ^{2}} d u

Take the constant out:

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

= 8 \cdot \int \frac{u}{( u ^{2} - 9 ) ^{2}} d u

Apply u-substitution

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

= 8 \cdot \int \frac{1}{2 v ^{2}} d v

Take the constant out:

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

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 8 \cdot \frac{1}{2} (- \frac{1}{v})

Substitute back

= 8 \cdot \frac{1}{2} (- \frac{1}{( 2 x + 3 ) ^{2} - 9})

Simplify

8 \cdot \frac{1}{2} (- \frac{1}{( 2 x + 3 ) ^{2} - 9}) : - \frac{1}{x ( x + 3 )}

= - \frac{1}{x ( x + 3 )}

Add a constant to the solution

= - \frac{1}{x ( x + 3 )} + C

x \to 0 lim (\frac{x}{∣ x ∣})

t - t^{- 1}

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

y^{^{''}} + 6 y^{^{'}} + 8 y = x^{2} e^{- x}

\frac{\partial}{\partial x} (7 e^{x^{2} l n (y)})

What is the integral of (2x+3)/((x^2+3x)^2) ?
The integral of (2x+3)/((x^2+3x)^2) is -1/(x(x+3))+C