What is the integral of (6x-5)/(sqrt(x^2-10x+9)) ?

The integral of (6x-5)/(sqrt(x^2-10x+9)) is 6sqrt(x^2-10x+9)+25ln(1/4 |sqrt(x^2-10x+9)+x-5|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (6x-5)/(sqrt(x^2-10x+9))

\int \frac{6 x - 5}{x ^{2} - 10 x + 9} d x

6 x^{2} - 10 x + 9 + 25 ln (\frac{1}{4} x^{2} - 10 x + 9 + x - 5) + C

Solution steps

\int \frac{6 x - 5}{x ^{2} - 10 x + 9} d x

Complete the square

x^{2} - 10 x + 9 : (x - 5)^{2} - 16

= \int \frac{6 x - 5}{( x - 5 ) ^{2} - 16} d x

Apply u-substitution

: \int \frac{6 u + 25}{u ^{2} - 16} d u

= \int \frac{6 u + 25}{u ^{2} - 16} d u

Expand

\frac{6 u + 25}{u ^{2} - 16} : \frac{6 u}{u ^{2} - 16} + \frac{25}{u ^{2} - 16}

= \int \frac{6 u}{u ^{2} - 16} + \frac{25}{u ^{2} - 16} d u

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{6 u}{u ^{2} - 16} d u + \int \frac{25}{u ^{2} - 16} d u

\int \frac{6 u}{u ^{2} - 16} d u = 6 u^{2} - 16

\int \frac{25}{u ^{2} - 16} d u = 25 ln (\frac{1}{4} u^{2} - 16 + u)

= 6 u^{2} - 16 + 25 ln (\frac{1}{4} u^{2} - 16 + u)

Substitute back

u = x - 5

= 6 (x - 5)^{2} - 16 + 25 ln (\frac{1}{4} (x - 5)^{2} - 16 + x - 5)

Simplify

6 (x - 5)^{2} - 16 + 25 ln (\frac{1}{4} (x - 5)^{2} - 16 + x - 5) : 6 x^{2} - 10 x + 9 + 25 ln (\frac{1}{4} x^{2} - 10 x + 9 + x - 5)

= 6 x^{2} - 10 x + 9 + 25 ln (\frac{1}{4} x^{2} - 10 x + 9 + x - 5)

Add a constant to the solution

= 6 x^{2} - 10 x + 9 + 25 ln (\frac{1}{4} x^{2} - 10 x + 9 + x - 5) + C

\int \frac{x ^{2} + 11 x - 30}{x ^{3} - 5 x ^{2} + 6 x} d x

\int \frac{x}{( 4 x ^{2} + 1 ) ^{5}} d x

\int 5 x x^{2} + 3 d x

\int - 8 x^{5} e^{x^{6}} d x

What is the integral of (6x-5)/(sqrt(x^2-10x+9)) ?
The integral of (6x-5)/(sqrt(x^2-10x+9)) is 6sqrt(x^2-10x+9)+25ln(1/4 |sqrt(x^2-10x+9)+x-5|)+C