What is the integral of 1/(x^2-6x+13) ?

The integral of 1/(x^2-6x+13) is 1/2 arctan((x-3)/2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x^2-6x+13)

\int \frac{1}{x ^{2} - 6 x + 13} d x

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

Solution steps

\int \frac{1}{x ^{2} - 6 x + 13} d x

Complete the square

x^{2} - 6 x + 13 : (x - 3)^{2} + 4

= \int \frac{1}{( x - 3 ) ^{2} + 4} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 4} d u

= \int \frac{1}{u ^{2} + 4} d u

Apply Integral Substitution

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

Substitute back

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

Add a constant to the solution

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

y^{^{''}} + 2 y^{^{'}} = 0, y (0) = - 4, y^{^{'}} (0) = 2

\frac{\partial}{\partial x} (x^{2} \cdot y)

\int x e^{(2 x)} d x

\int_{0}^{5} \frac{3 x}{( x - 6 ) ^{2}} d x

\int_{e}^{\infty} x^{1} ln (x) d x

What is the integral of 1/(x^2-6x+13) ?
The integral of 1/(x^2-6x+13) is 1/2 arctan((x-3)/2)+C