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

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

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

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

3 (\frac{x ^{2}}{x + 2} - \frac{4}{x + 2} + ln ∣ x - 2 ∣ - ln ∣ x + 2 ∣) + C

Solution steps

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

Factor

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

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

Cancel the common factor:

x

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

Apply u-substitution

: \int \frac{( u - 2 ) ^{2}}{u ( u - 4 )} d u

= 3 \cdot \int \frac{( u - 2 ) ^{2}}{u ( u - 4 )} d u

Apply u-substitution

: \int \frac{( v + 2 ) ^{2}}{v ( v + 4 )} d v

= 3 \cdot \int \frac{( v + 2 ) ^{2}}{v ( v + 4 )} d v

Apply Integration By Parts

: \frac{( v + 2 ) ^{2}}{v + 4} - \int \frac{- 8 v - 16}{v ( v + 4 ) ^{2}} d v

= 3 (\frac{( v + 2 ) ^{2}}{v + 4} - \int \frac{- 8 v - 16}{v ( v + 4 ) ^{2}} d v)

\int \frac{- 8 v - 16}{v ( v + 4 ) ^{2}} d v = \frac{4}{v + 4} - ln ∣ v ∣ + ln ∣ v + 4 ∣

= 3 (\frac{( v + 2 ) ^{2}}{v + 4} - (\frac{4}{v + 4} - ln ∣ v ∣ + ln ∣ v + 4 ∣))

Substitute back

= 3 (\frac{( x + 2 - 4 + 2 ) ^{2}}{x + 2 - 4 + 4} - (\frac{4}{x + 2 - 4 + 4} - ln ∣ x + 2 - 4 ∣ + ln ∣ x + 2 - 4 + 4 ∣))

Simplify

3 (\frac{( x + 2 - 4 + 2 ) ^{2}}{x + 2 - 4 + 4} - (\frac{4}{x + 2 - 4 + 4} - ln ∣ x + 2 - 4 ∣ + ln ∣ x + 2 - 4 + 4 ∣)) : 3 (\frac{x ^{2}}{x + 2} - \frac{4}{x + 2} + ln ∣ x - 2 ∣ - ln ∣ x + 2 ∣)

= 3 (\frac{x ^{2}}{x + 2} - \frac{4}{x + 2} + ln ∣ x - 2 ∣ - ln ∣ x + 2 ∣)

Add a constant to the solution

= 3 (\frac{x ^{2}}{x + 2} - \frac{4}{x + 2} + ln ∣ x - 2 ∣ - ln ∣ x + 2 ∣) + C

(6 n - 1) (- 2 x)^{n}

\frac{d y}{d x} = k y^{2}

f (x) = 2 x, (8, 4)

\int y^{2} e^{x y} + cos (x) d x

\frac{\partial}{\partial y} (1 - y^{3})

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