What is the integral of (8x^2+6)/((x-5)^3) ?

The integral of (8x^2+6)/((x-5)^3) is 8ln|x-5|-(80)/(x-5)-(103)/((x-5)^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (8x^2+6)/((x-5)^3)

\int \frac{8 x ^{2} + 6}{( x - 5 ) ^{3}} d x

8 ln ∣ x - 5 ∣ - \frac{80}{x - 5} - \frac{103}{( x - 5 ) ^{2}} + C

Solution steps

\int \frac{8 x ^{2} + 6}{( x - 5 ) ^{3}} d x

Apply u-substitution

: \int \frac{8 u ^{2} + 80 u + 206}{u ^{3}} d u

= \int \frac{8 u ^{2} + 80 u + 206}{u ^{3}} d u

Expand

\frac{8 u ^{2} + 80 u + 206}{u ^{3}} : \frac{8}{u} + \frac{80}{u ^{2}} + \frac{206}{u ^{3}}

= \int \frac{8}{u} + \frac{80}{u ^{2}} + \frac{206}{u ^{3}} 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{8}{u} d u + \int \frac{80}{u ^{2}} d u + \int \frac{206}{u ^{3}} d u

\int \frac{8}{u} d u = 8 ln ∣ u ∣

\int \frac{80}{u ^{2}} d u = - \frac{80}{u}

\int \frac{206}{u ^{3}} d u = - \frac{103}{u ^{2}}

= 8 ln ∣ u ∣ - \frac{80}{u} - \frac{103}{u ^{2}}

Substitute back

u = x - 5

= 8 ln ∣ x - 5 ∣ - \frac{80}{x - 5} - \frac{103}{( x - 5 ) ^{2}}

Add a constant to the solution

= 8 ln ∣ x - 5 ∣ - \frac{80}{x - 5} - \frac{103}{( x - 5 ) ^{2}} + C

\frac{d}{d x} (arctan (x + 1) + 2 x - y)

y^{^{''}} + 25 y = cos (2 x)

\int \frac{( x + 7 ) ^{5}}{x} d x

\frac{d y}{d x} = \frac{- y ^{2}}{1 + y ^{2}} \cdot e^{- x}

\frac{\partial}{\partial z} ((x + z) e^{x - y})

What is the integral of (8x^2+6)/((x-5)^3) ?
The integral of (8x^2+6)/((x-5)^3) is 8ln|x-5|-(80)/(x-5)-(103)/((x-5)^2)+C