What is the integral of x/(sqrt(x^2-20x+84)) ?

The integral of x/(sqrt(x^2-20x+84)) is sqrt(x^2-20x+84)+10ln(1/4 |sqrt(x^2-20x+84)+x-10|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x/(sqrt(x^2-20x+84))

\int \frac{x}{x ^{2} - 20 x + 84} d x

x^{2} - 20 x + 84 + 10 ln (\frac{1}{4} x^{2} - 20 x + 84 + x - 10) + C

Solution steps

\int \frac{x}{x ^{2} - 20 x + 84} d x

Complete the square

x^{2} - 20 x + 84 : (x - 10)^{2} - 16

= \int \frac{x}{( x - 10 ) ^{2} - 16} d x

Apply u-substitution

: \int \frac{u + 10}{u ^{2} - 16} d u

= \int \frac{u + 10}{u ^{2} - 16} d u

Expand

\frac{u + 10}{u ^{2} - 16} : \frac{u}{u ^{2} - 16} + \frac{10}{u ^{2} - 16}

= \int \frac{u}{u ^{2} - 16} + \frac{10}{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{u}{u ^{2} - 16} d u + \int \frac{10}{u ^{2} - 16} d u

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

\int \frac{10}{u ^{2} - 16} d u = 10 ln (\frac{1}{4} u^{2} - 16 + u)

= u^{2} - 16 + 10 ln (\frac{1}{4} u^{2} - 16 + u)

Substitute back

u = x - 10

= (x - 10)^{2} - 16 + 10 ln (\frac{1}{4} (x - 10)^{2} - 16 + x - 10)

Simplify

(x - 10)^{2} - 16 + 10 ln (\frac{1}{4} (x - 10)^{2} - 16 + x - 10) : x^{2} - 20 x + 84 + 10 ln (\frac{1}{4} x^{2} - 20 x + 84 + x - 10)

= x^{2} - 20 x + 84 + 10 ln (\frac{1}{4} x^{2} - 20 x + 84 + x - 10)

Add a constant to the solution

= x^{2} - 20 x + 84 + 10 ln (\frac{1}{4} x^{2} - 20 x + 84 + x - 10) + C

\int \frac{1}{x ( 9 ln ( x ) + 8 )} d x

\int (6 z + 2)^{\frac{1}{2}} d z

\int 4 x ln (9 x) d x

\int 9 x sin (3 x) d x

\int \frac{( sin ^{3} ( x ) )}{cos ( x )} d x

What is the integral of x/(sqrt(x^2-20x+84)) ?
The integral of x/(sqrt(x^2-20x+84)) is sqrt(x^2-20x+84)+10ln(1/4 |sqrt(x^2-20x+84)+x-10|)+C