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

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

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

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

- 4 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 - \frac{x}{2}) + C

Solution steps

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

Factor

x^{2} - 4 : - (- x^{2} + 4)

= 2 \cdot \int \frac{x ^{2}}{- ( - x ^{2} + 4 )} d x

Take the constant out:

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

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

Apply Integral Substitution

= 2 (- \int \frac{2 u ^{2}}{- u ^{2} + 1} d u)

Take the constant out:

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

= 2 (- 2 \cdot \int \frac{u ^{2}}{- u ^{2} + 1} d u)

\frac{u ^{2}}{- u ^{2} + 1} = \frac{1}{- u ^{2} + 1} - 1

= 2 (- 2 \cdot \int \frac{1}{- u ^{2} + 1} - 1 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 (- 2 (\int \frac{1}{- u ^{2} + 1} d u - \int 1 d u))

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

\int 1 d u = u

= 2 (- 2 (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2} - u))

Substitute back

u = \frac{x}{2}

= 2 (- 2 (\frac{ln \frac{x}{2} + 1}{2} - \frac{ln \frac{x}{2} - 1}{2} - \frac{x}{2}))

Simplify

2 (- 2 (\frac{ln \frac{x}{2} + 1}{2} - \frac{ln \frac{x}{2} - 1}{2} - \frac{x}{2})) : - 4 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 - \frac{x}{2})

= - 4 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 - \frac{x}{2})

Add a constant to the solution

= - 4 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 - \frac{x}{2}) + C

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