What is the integral of 1/(4y^2+12y+9) ?

The integral of 1/(4y^2+12y+9) is -1/(2(2y+3))+C

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integral of 1/(4y^2+12y+9)

\int \frac{1}{4 y ^{2} + 12 y + 9} d y

- \frac{1}{2 ( 2 y + 3 )} + C

Solution steps

\int \frac{1}{4 y ^{2} + 12 y + 9} d y

Complete the square

4 y^{2} + 12 y + 9 : 4 (y + \frac{3}{2})^{2}

= \int \frac{1}{4 ( y + \frac{3}{2} ) ^{2}} d y

Apply u-substitution

: \int \frac{1}{4 u ^{2}} d u

= \int \frac{1}{4 u ^{2}} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= \frac{1}{4} (- \frac{1}{u})

Substitute back

u = y + \frac{3}{2}

= \frac{1}{4} (- \frac{1}{y + \frac{3}{2}})

Simplify

\frac{1}{4} (- \frac{1}{y + \frac{3}{2}}) : - \frac{1}{2 ( 2 y + 3 )}

= - \frac{1}{2 ( 2 y + 3 )}

Add a constant to the solution

= - \frac{1}{2 ( 2 y + 3 )} + C

f (t) = \frac{t ^{3}}{3}

\frac{d}{d y} (\frac{d y}{y})

\int 16 - 25 x^{2} d x

(4 x y - 18 x y) d x + (2 x^{2} - 9 x^{2} + 1) d y = 0

\frac{d}{d x} (\frac{4}{5})

What is the integral of 1/(4y^2+12y+9) ?
The integral of 1/(4y^2+12y+9) is -1/(2(2y+3))+C