What is the integral of-1/(5tsqrt(25t^2-25)) ?

The integral of-1/(5tsqrt(25t^2-25)) is -1/25 arctan(sqrt(t^2-1))+C

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integral of-1/(5tsqrt(25t^2-25))

\int - \frac{1}{5 t 25 t ^{2} - 25} d t

- \frac{1}{25} arctan (t^{2} - 1) + C

Solution steps

\int - \frac{1}{5 t 25 t ^{2} - 25} d t

Take the constant out:

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

= - \frac{1}{5} \cdot \int \frac{1}{t 25 t ^{2} - 25} d t

Apply Integration By Parts

: \frac{1}{5 t ^{2} - 1} - \int - \frac{2 t ^{2} - 1}{5 t ( t ^{2} - 1 ) t ^{2} - 1} d t

= - \frac{1}{5} (\frac{1}{5 t ^{2} - 1} - \int - \frac{2 t ^{2} - 1}{5 t ( t ^{2} - 1 ) t ^{2} - 1} d t)

\int - \frac{2 t ^{2} - 1}{5 t ( t ^{2} - 1 ) t ^{2} - 1} d t = - \frac{1}{5} (- \frac{1}{t ^{2} - 1} + arctan (t^{2} - 1))

= - \frac{1}{5} (\frac{1}{5 t ^{2} - 1} - (- \frac{1}{5} (- \frac{1}{t ^{2} - 1} + arctan (t^{2} - 1))))

Simplify

- \frac{1}{5} (\frac{1}{5 t ^{2} - 1} - (- \frac{1}{5} (- \frac{1}{t ^{2} - 1} + arctan (t^{2} - 1)))) : - \frac{1}{25} arctan (t^{2} - 1)

= - \frac{1}{25} arctan (t^{2} - 1)

Add a constant to the solution

= - \frac{1}{25} arctan (t^{2} - 1) + C

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What is the integral of-1/(5tsqrt(25t^2-25)) ?
The integral of-1/(5tsqrt(25t^2-25)) is -1/25 arctan(sqrt(t^2-1))+C