What is the integral of (x^2)/(x-3) ?

The integral of (x^2)/(x-3) is ((x-3)^2)/2+6(x-3)+9ln|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 (x^2)/(x-3)

\int \frac{x ^{2}}{x - 3} d x

\frac{( x - 3 ) ^{2}}{2} + 6 (x - 3) + 9 ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{x ^{2}}{x - 3} d x

Apply u-substitution

: \int \frac{( u + 3 ) ^{2}}{u} d u

= \int \frac{( u + 3 ) ^{2}}{u} d u

Expand

\frac{( u + 3 ) ^{2}}{u} : u + 6 + \frac{9}{u}

= \int u + 6 + \frac{9}{u} 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 u d u + \int 6 d u + \int \frac{9}{u} d u

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

\int 6 d u = 6 u

\int \frac{9}{u} d u = 9 ln ∣ u ∣

= \frac{u ^{2}}{2} + 6 u + 9 ln ∣ u ∣

Substitute back

u = x - 3

= \frac{( x - 3 ) ^{2}}{2} + 6 (x - 3) + 9 ln ∣ x - 3 ∣

Add a constant to the solution

= \frac{( x - 3 ) ^{2}}{2} + 6 (x - 3) + 9 ln ∣ x - 3 ∣ + C

\int \frac{6 x}{4 x ^{2} + 4 x + 10} d x

\frac{d}{d x} (e^{s i n (x - 1)})

\int_{2}^{5} x^{6} d x

x y^{^{'}} - 2 y^{^{'}} - (x - 2) cos (x) + y = 0

\int - 11 sec^{2} (x) d x

What is the integral of (x^2)/(x-3) ?
The integral of (x^2)/(x-3) is ((x-3)^2)/2+6(x-3)+9ln|x-3|+C