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

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

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integral of 4/(4-t^2)

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

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

Solution steps

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

Take the constant out:

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

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

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{t}{2}

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

Simplify

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

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

Add a constant to the solution

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

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