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

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

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integral of 1/(x^2-4x+13)

\int \frac{1}{x ^{2} - 4 x + 13} d x

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

Solution steps

\int \frac{1}{x ^{2} - 4 x + 13} d x

Complete the square

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

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

Apply u-substitution

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

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

Apply Integral Substitution

= \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

= \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)

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

Substitute back

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

Add a constant to the solution

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

\frac{d}{d x} (e^{x} sin (x) cos^{3} (x))

\int (\frac{x ^{2}}{5 + 2 x ^{3}}) d x

x \to 0 lim (\frac{5 x ^{n} - 3 x ^{n - 1} + 4 x ^{n - 2}}{2 x ^{n} - 6 x ^{n - 2}})

f (x) = 5 x + 3

\frac{\partial}{\partial x} (- 4 y^{4} sin (4 x))

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