What is the integral of 1/(x^2+8x+52) ?

The integral of 1/(x^2+8x+52) is 1/6 arctan((x+4)/6)+C

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integral of 1/(x^2+8x+52)

\int \frac{1}{x ^{2} + 8 x + 52} d x

\frac{1}{6} arctan (\frac{x + 4}{6}) + C

Solution steps

\int \frac{1}{x ^{2} + 8 x + 52} d x

Complete the square

x^{2} + 8 x + 52 : (x + 4)^{2} + 36

= \int \frac{1}{( x + 4 ) ^{2} + 36} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 36} d u

= \int \frac{1}{u ^{2} + 36} d u

Apply Integral Substitution

= \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

= \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)

= \frac{1}{6} arctan (v)

Substitute back

= \frac{1}{6} arctan (\frac{x + 4}{6})

Add a constant to the solution

= \frac{1}{6} arctan (\frac{x + 4}{6}) + C

x^{2} y^{^{'}} = y^{2} + 3 x y

\frac{d}{d x} (1 + x^{- 1})

(9, 2.6), (8, 2.2)

\int sinh (2 t) d t

\int_{0}^{1} 2 x + 1 d x

What is the integral of 1/(x^2+8x+52) ?
The integral of 1/(x^2+8x+52) is 1/6 arctan((x+4)/6)+C