What is the integral of 4/(4x^2+20x+34) ?

The integral of 4/(4x^2+20x+34) is 2/3 arctan(2/3 x+5/3)+C

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integral of 4/(4x^2+20x+34)

\int \frac{4}{4 x ^{2} + 20 x + 34} d x

\frac{2}{3} arctan (\frac{2}{3} x + \frac{5}{3}) + C

Solution steps

\int \frac{4}{4 x ^{2} + 20 x + 34} 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} + 20 x + 34} d x

Complete the square

4 x^{2} + 20 x + 34 : 4 (x + \frac{5}{2})^{2} + 9

= 4 \cdot \int \frac{1}{4 ( x + \frac{5}{2} ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{1}{4 u ^{2} + 9} d u

= 4 \cdot \int \frac{1}{4 u ^{2} + 9} d u

Apply Integral Substitution

= 4 \cdot \int \frac{1}{6 ( 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}{6} \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}{6} arctan (v)

Substitute back

= 4 \cdot \frac{1}{6} arctan (\frac{2}{3} (x + \frac{5}{2}))

Simplify

4 \cdot \frac{1}{6} arctan (\frac{2}{3} (x + \frac{5}{2})) : \frac{2}{3} arctan (\frac{2}{3} x + \frac{5}{3})

= \frac{2}{3} arctan (\frac{2}{3} x + \frac{5}{3})

Add a constant to the solution

= \frac{2}{3} arctan (\frac{2}{3} x + \frac{5}{3}) + C

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What is the integral of 4/(4x^2+20x+34) ?
The integral of 4/(4x^2+20x+34) is 2/3 arctan(2/3 x+5/3)+C