What is the integral of 1/(x^2-6x+25) ?

The integral of 1/(x^2-6x+25) is 1/4 arctan((x-3)/4)+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+25)

\int \frac{1}{x ^{2} - 6 x + 25} d x

\frac{1}{4} arctan (\frac{x - 3}{4}) + C

Solution steps

\int \frac{1}{x ^{2} - 6 x + 25} d x

Complete the square

x^{2} - 6 x + 25 : (x - 3)^{2} + 16

= \int \frac{1}{( x - 3 ) ^{2} + 16} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 16} d u

= \int \frac{1}{u ^{2} + 16} d u

Apply Integral Substitution

= \int \frac{1}{4 ( 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}{4} \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}{4} arctan (v)

Substitute back

= \frac{1}{4} arctan (\frac{x - 3}{4})

Add a constant to the solution

= \frac{1}{4} arctan (\frac{x - 3}{4}) + C

\int \frac{x}{x ^{2} - 4 x - 5} d x

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

\int (x^{2} + x + 1) ln (x) d x

\int - 2 x e^{2 x} d x

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

What is the integral of 1/(x^2-6x+25) ?
The integral of 1/(x^2-6x+25) is 1/4 arctan((x-3)/4)+C