What is the integral of 2/(x^2-4x+13) ?

The integral of 2/(x^2-4x+13) is 2/3 arctan(1/3 (x-2))+C

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integral of 2/(x^2-4x+13)

\int \frac{2}{x ^{2} - 4 x + 13} d x

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

Solution steps

\int \frac{2}{x ^{2} - 4 x + 13} 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} - 4 x + 13} d x

Complete the square

x^{2} - 4 x + 13 : (x - 2)^{2} + 9

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

Apply u-substitution

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

= 2 \cdot \frac{1}{3} arctan (v)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

n = 0 \sum \infty \frac{1}{n ^{4}}

\frac{d y}{d x} + 3 x^{2} y = 18 x^{2}, y (0) = 8

38 (4 e^{x^{2}} x^{3} + 6 e^{x^{2}} x)

n = 1 \sum \infty 0. 3^{n}

\frac{\partial}{\partial x} (- 3 x y^{2} - 2 y cos (x))

What is the integral of 2/(x^2-4x+13) ?
The integral of 2/(x^2-4x+13) is 2/3 arctan(1/3 (x-2))+C