What is the integral of ((3x^4))/((x^2-2x)) ?

The integral of ((3x^4))/((x^2-2x)) is 9x^2+24ln|x-2|+(x-2)^3-36+C

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integral of ((3x^4))/((x^2-2x))

\int \frac{( 3 x ^{4} )}{( x ^{2} - 2 x )} d x

9 x^{2} + 24 ln ∣ x - 2 ∣ + (x - 2)^{3} - 36 + C

Solution steps

\int \frac{3 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

= 3 \cdot \int \frac{x ^{4}}{x ^{2} - 2 x} d x

Simplify

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

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

Apply u-substitution

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

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

Expand

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

= 3 \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

= 3 (\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 ∣

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

Substitute back

u = x - 2

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

Simplify

3 (\frac{( x - 2 ) ^{3}}{3} + 3 (x - 2)^{2} + 12 (x - 2) + 8 ln ∣ x - 2 ∣) : 9 x^{2} + 24 ln ∣ x - 2 ∣ + (x - 2)^{3} - 36

= 9 x^{2} + 24 ln ∣ x - 2 ∣ + (x - 2)^{3} - 36

Add a constant to the solution

= 9 x^{2} + 24 ln ∣ x - 2 ∣ + (x - 2)^{3} - 36 + C

x y^{^{'}} + y = 11 x

\frac{d}{d x} (x (6 - x)^{\frac{2}{3}})

\int \frac{x ^{2} - 4}{x ^{3}} d x

\frac{d}{d x} (x^{2} e^{2} + 2^{y}) = 5

4 x^{4} - 4 x^{2}, x^{2}

What is the integral of ((3x^4))/((x^2-2x)) ?
The integral of ((3x^4))/((x^2-2x)) is 9x^2+24ln|x-2|+(x-2)^3-36+C