What is the integral of 4/(4x^2+4x+65) ?

The integral of 4/(4x^2+4x+65) is 1/4 arctan(1/8 (2x+1))+C

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integral of 4/(4x^2+4x+65)

\int \frac{4}{4 x ^{2} + 4 x + 65} d x

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

Solution steps

\int \frac{4}{4 x ^{2} + 4 x + 65} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{4 x ^{2} + 4 x + 65} d x

Complete the square

4 x^{2} + 4 x + 65 : 4 (x + \frac{1}{2})^{2} + 64

= 4 \cdot \int \frac{1}{4 ( x + \frac{1}{2} ) ^{2} + 64} d x

Apply u-substitution

: \int \frac{1}{4 u ^{2} + 64} d u

= 4 \cdot \int \frac{1}{4 u ^{2} + 64} d u

Apply Integral Substitution

= 4 \cdot \int \frac{1}{16 ( v ^{2} + 1 )} d v

Take the constant out:

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

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

Use the common integral:

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

= 4 \cdot \frac{1}{16} arctan (v)

Substitute back

= 4 \cdot \frac{1}{16} arctan (\frac{2}{8} (x + \frac{1}{2}))

Simplify

4 \cdot \frac{1}{16} arctan (\frac{2}{8} (x + \frac{1}{2})) : \frac{1}{4} arctan (\frac{1}{8} (2 x + 1))

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

Add a constant to the solution

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

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What is the integral of 4/(4x^2+4x+65) ?
The integral of 4/(4x^2+4x+65) is 1/4 arctan(1/8 (2x+1))+C