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

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

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

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

- ln \frac{x}{2} + 1 + ln \frac{x}{2} - 1 + x + C

Solution steps

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

Factor

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

= \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

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

Apply Integral Substitution

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

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

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

Substitute back

u = \frac{x}{2}

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

Simplify

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

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

Add a constant to the solution

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

\int x (6 + 10 x) d x

\int 13 sec^{- 2} (x t a) n^{3} x d x

y^{^{''}} + y^{^{'}} + y = 0.5

f^{^{'}} (x) = 5

\frac{d}{d x} (\frac{( e ^{x} )}{ln ( x )})

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