What is the integral of 1/(4x^2+12x+25) ?

The integral of 1/(4x^2+12x+25) is 1/8 arctan(1/4 (2x+3))+C

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integral of 1/(4x^2+12x+25)

\int \frac{1}{4 x ^{2} + 12 x + 25} d x

\frac{1}{8} arctan (\frac{1}{4} (2 x + 3)) + C

Solution steps

\int \frac{1}{4 x ^{2} + 12 x + 25} d x

Complete the square

4 x^{2} + 12 x + 25 : 4 (x + \frac{3}{2})^{2} + 16

= \int \frac{1}{4 ( x + \frac{3}{2} ) ^{2} + 16} d x

Apply u-substitution

= \int \frac{1}{4 u ^{2} + 16} d u

Apply Integral Substitution

= \int \frac{1}{8 ( v ^{2} + 1 )} d v

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{8} arctan (v)

Substitute back

= \frac{1}{8} arctan (\frac{2}{4} (x + \frac{3}{2}))

Simplify

\frac{1}{8} arctan (\frac{2}{4} (x + \frac{3}{2})) : \frac{1}{8} arctan (\frac{1}{4} (2 x + 3))

= \frac{1}{8} arctan (\frac{1}{4} (2 x + 3))

Add a constant to the solution

= \frac{1}{8} arctan (\frac{1}{4} (2 x + 3)) + C

What is the integral of 1/(4x^2+12x+25) ?
The integral of 1/(4x^2+12x+25) is 1/8 arctan(1/4 (2x+3))+C