What is the integral of (2x^4)/(x^2-2x) ?

The integral of (2x^4)/(x^2-2x) is 2(3x^2+8ln|x-2|+1/3 (x-2)^3-12)+C

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integral of (2x^4)/(x^2-2x)

\int \frac{2 x ^{4}}{x ^{2} - 2 x} d x

2 (3 x^{2} + 8 ln ∣ x - 2 ∣ + \frac{1}{3} (x - 2)^{3} - 12) + C

Solution steps

\int \frac{2 x ^{4}}{x ^{2} - 2 x} 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 ^{4}}{x ^{2} - 2 x} d x

Simplify

\frac{x ^{4}}{x ^{2} - 2 x} : \frac{x ^{3}}{x - 2}

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

Apply u-substitution

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

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

Expand

\frac{( u + 2 ) ^{3}}{u} : u^{2} + 6 u + 12 + \frac{8}{u}

= 2 \cdot \int u^{2} + 6 u + 12 + \frac{8}{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^{2} d u + \int 6 u d u + \int 12 d u + \int \frac{8}{u} d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 6 u d u = 3 u^{2}

\int 12 d u = 12 u

\int \frac{8}{u} d u = 8 ln ∣ u ∣

= 2 (\frac{u ^{3}}{3} + 3 u^{2} + 12 u + 8 ln ∣ u ∣)

Substitute back

u = x - 2

= 2 (\frac{( x - 2 ) ^{3}}{3} + 3 (x - 2)^{2} + 12 (x - 2) + 8 ln ∣ x - 2 ∣)

Simplify

2 (\frac{( x - 2 ) ^{3}}{3} + 3 (x - 2)^{2} + 12 (x - 2) + 8 ln ∣ x - 2 ∣) : 2 (3 x^{2} + 8 ln ∣ x - 2 ∣ + \frac{1}{3} (x - 2)^{3} - 12)

= 2 (3 x^{2} + 8 ln ∣ x - 2 ∣ + \frac{1}{3} (x - 2)^{3} - 12)

Add a constant to the solution

= 2 (3 x^{2} + 8 ln ∣ x - 2 ∣ + \frac{1}{3} (x - 2)^{3} - 12) + C

\int_{0}^{1} x 1 + 4 x^{2} d x

\frac{\partial}{\partial x} (x^{2} - 5 x y)

f (x) = x^{1.7}

x \to - 4 lim (\frac{x ^{2} + 9}{x + 2})

\int 16 e^{u} d u

What is the integral of (2x^4)/(x^2-2x) ?
The integral of (2x^4)/(x^2-2x) is 2(3x^2+8ln|x-2|+1/3 (x-2)^3-12)+C