What is the integral of 4/(tsqrt(t^2-1)) ?

The integral of 4/(tsqrt(t^2-1)) is 4arctan(sqrt(t^2-1))+C

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integral of 4/(tsqrt(t^2-1))

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

4 arctan (t^{2} - 1) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2} + 1} d u

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

Use the common integral:

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

= 4 arctan (u)

Substitute back

u = t^{2} - 1

= 4 arctan (t^{2} - 1)

Add a constant to the solution

= 4 arctan (t^{2} - 1) + C

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What is the integral of 4/(tsqrt(t^2-1)) ?
The integral of 4/(tsqrt(t^2-1)) is 4arctan(sqrt(t^2-1))+C