What is the integral of (2x-5)/(x^2-6x+10) ?

The integral of (2x-5)/(x^2-6x+10) is ln|x^2-6x+10|+arctan(x-3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x-5)/(x^2-6x+10)

\int \frac{2 x - 5}{x ^{2} - 6 x + 10} d x

ln x^{2} - 6 x + 10 + arctan (x - 3) + C

Solution steps

\int \frac{2 x - 5}{x ^{2} - 6 x + 10} d x

Complete the square

x^{2} - 6 x + 10 : (x - 3)^{2} + 1

= \int \frac{2 x - 5}{( x - 3 ) ^{2} + 1} d x

Apply u-substitution

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

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

Expand

\frac{2 u + 1}{u ^{2} + 1} : \frac{2 u}{u ^{2} + 1} + \frac{1}{u ^{2} + 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

= ln u^{2} + 1 + arctan (u)

Substitute back

u = x - 3

= ln (x - 3)^{2} + 1 + arctan (x - 3)

Expand

(x - 3)^{2} + 1 : x^{2} - 6 x + 10

= ln x^{2} - 6 x + 10 + arctan (x - 3)

Add a constant to the solution

= ln x^{2} - 6 x + 10 + arctan (x - 3) + C

\int \frac{sin ( x )}{cos ( x ) cos ( 2 x )} d x

\int (4 + cos (x))^{2} d x

\int \frac{1}{cos ^{2} ( 2 x - 3 )} d x

\int 9 (9 x + 2)^{4} d x

\int (2 sin (2 x)) d x

What is the integral of (2x-5)/(x^2-6x+10) ?
The integral of (2x-5)/(x^2-6x+10) is ln|x^2-6x+10|+arctan(x-3)+C