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

The integral of (2x^2)/(x-3) is x^2+6x+18ln|x-3|-27+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x^2)/(x-3)

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

x^{2} + 6 x + 18 ln ∣ x - 3 ∣ - 27 + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

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

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

Expand

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

= 2 \cdot \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

= 2 (\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 ∣

= 2 (\frac{u ^{2}}{2} + 6 u + 9 ln ∣ u ∣)

Substitute back

u = x - 3

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

Simplify

2 (\frac{( x - 3 ) ^{2}}{2} + 6 (x - 3) + 9 ln ∣ x - 3 ∣) : x^{2} + 6 x + 18 ln ∣ x - 3 ∣ - 27

= x^{2} + 6 x + 18 ln ∣ x - 3 ∣ - 27

Add a constant to the solution

= x^{2} + 6 x + 18 ln ∣ x - 3 ∣ - 27 + C

\int \frac{1 - sin ( x )}{1 + cos ( x )} d x

\int e^{2 x} csc (e^{2 x}) d x

\int sin^{5} (3 x) cos^{2} (3 x) d x

\int 1 0^{x} sin (1 0^{x}) d x

\int cos^{9} (5 x) sin (5 x) d x

What is the integral of (2x^2)/(x-3) ?
The integral of (2x^2)/(x-3) is x^2+6x+18ln|x-3|-27+C