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

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

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integral of 4/(4-x^2)

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

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

Solution steps

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

Take the constant out:

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

= 4 \cdot \int \frac{1}{4 - x ^{2}} d x

Apply Integral Substitution

= 4 \cdot \int \frac{1}{2 ( - u ^{2} + 1 )} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{- u ^{2} + 1} d u

Use the common integral:

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

= 4 \cdot \frac{1}{2} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = \frac{x}{2}

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

Simplify

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

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

Add a constant to the solution

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

- \frac{6 x}{x ^{2} + 1}, at x = 1

f (x) = 9 x^{2}

\int \frac{x}{1 - 5 x ^{2}} d x

\frac{d}{d x} (\frac{1}{x ^{13}})

\int 4 tan^{3} (x) sec (x) d x

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