What is the integral of 3/(x^2+4x+14) ?

The integral of 3/(x^2+4x+14) is 3/(sqrt(10))arctan(1/(sqrt(10))(x+2))+C

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integral of 3/(x^2+4x+14)

\int \frac{3}{x ^{2} + 4 x + 14} d x

\frac{3}{10} arctan (\frac{1}{10} (x + 2)) + C

Solution steps

\int \frac{3}{x ^{2} + 4 x + 14} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{x ^{2} + 4 x + 14} d x

Complete the square

x^{2} + 4 x + 14 : (x + 2)^{2} + 10

= 3 \cdot \int \frac{1}{( x + 2 ) ^{2} + 10} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 10} d u

= 3 \cdot \int \frac{1}{u ^{2} + 10} d u

Apply Integral Substitution

= 3 \cdot \int \frac{1}{10 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 3 \cdot \frac{1}{10} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 3 \cdot \frac{1}{10} arctan (v)

Substitute back

= 3 \cdot \frac{1}{10} arctan (\frac{x + 2}{10})

Simplify

3 \cdot \frac{1}{10} arctan (\frac{x + 2}{10}) : \frac{3}{10} arctan (\frac{1}{10} (x + 2))

= \frac{3}{10} arctan (\frac{1}{10} (x + 2))

Add a constant to the solution

= \frac{3}{10} arctan (\frac{1}{10} (x + 2)) + C

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What is the integral of 3/(x^2+4x+14) ?
The integral of 3/(x^2+4x+14) is 3/(sqrt(10))arctan(1/(sqrt(10))(x+2))+C