What is the integral of 1/(x^2-2x+10) ?

The integral of 1/(x^2-2x+10) is 1/3 arctan((x-1)/3)+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^2-2x+10)

\int \frac{1}{x ^{2} - 2 x + 10} d x

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

Solution steps

\int \frac{1}{x ^{2} - 2 x + 10} d x

Complete the square

x^{2} - 2 x + 10 : (x - 1)^{2} + 9

= \int \frac{1}{( x - 1 ) ^{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 - 1}{3})

Add a constant to the solution

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

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

(- x^{3} + 4 x^{2})^{3}

\int 15 t^{2} d t

y^{^{'}} = \frac{1}{y - x}

\int (12 x + 2) 6 x^{2} + 2 x d x

What is the integral of 1/(x^2-2x+10) ?
The integral of 1/(x^2-2x+10) is 1/3 arctan((x-1)/3)+C