What is the integral of 2/(x^2+8x+17) ?

The integral of 2/(x^2+8x+17) is 2arctan(x+4)+C

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integral of 2/(x^2+8x+17)

\int \frac{2}{x ^{2} + 8 x + 17} d x

2 arctan (x + 4) + C

Solution steps

\int \frac{2}{x ^{2} + 8 x + 17} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{x ^{2} + 8 x + 17} d x

Complete the square

x^{2} + 8 x + 17 : (x + 4)^{2} + 1

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

Apply u-substitution

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

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

Use the common integral:

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

= 2 arctan (u)

Substitute back

u = x + 4

= 2 arctan (x + 4)

Add a constant to the solution

= 2 arctan (x + 4) + C

\int 42 cos^{2} (97 x) d x

y^{^{''}} - 0.2 y^{^{'}} + 0.01 y = 0, y (0) = 15, y^{^{'}} (0) = b

x d y - (2 x + 1) e^{- y} d x = 0

\int \frac{1}{( 3 x - 1 )} d x

\frac{\partial}{\partial x} (2 cos (3 x))

What is the integral of 2/(x^2+8x+17) ?
The integral of 2/(x^2+8x+17) is 2arctan(x+4)+C