What is the integral of 1/(x^2+8x+32) ?

The integral of 1/(x^2+8x+32) is 1/4 arctan((x+4)/4)+C

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integral of 1/(x^2+8x+32)

\int \frac{1}{x ^{2} + 8 x + 32} d x

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

Solution steps

\int \frac{1}{x ^{2} + 8 x + 32} d x

Complete the square

x^{2} + 8 x + 32 : (x + 4)^{2} + 16

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

Apply u-substitution

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

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

Apply Integral Substitution

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

Substitute back

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

Add a constant to the solution

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

x \to 7 - lim (\frac{x}{x ^{2} - 49})

x + 1, 9 - x^{2}, - 1, 2

y^{^{'}} = (x + y)^{2} - 1, y (1) = 6

x \to 1 - lim (x^{3})

sin (7 ln (x))

What is the integral of 1/(x^2+8x+32) ?
The integral of 1/(x^2+8x+32) is 1/4 arctan((x+4)/4)+C