What is the integral of 8/(5tsqrt(25t^2-25)) ?

The integral of 8/(5tsqrt(25t^2-25)) is 8/25 arctan(sqrt(t^2-1))+C

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integral of 8/(5tsqrt(25t^2-25))

\int \frac{8}{5 t 25 t ^{2} - 25} d t

\frac{8}{25} arctan (t^{2} - 1) + C

Solution steps

\int \frac{8}{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{8}{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{8}{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{8}{5} (\frac{1}{5 t ^{2} - 1} - (- \frac{1}{5} (- \frac{1}{t ^{2} - 1} + arctan (t^{2} - 1))))

Simplify

\frac{8}{5} (\frac{1}{5 t ^{2} - 1} - (- \frac{1}{5} (- \frac{1}{t ^{2} - 1} + arctan (t^{2} - 1)))) : \frac{8}{25} arctan (t^{2} - 1)

= \frac{8}{25} arctan (t^{2} - 1)

Add a constant to the solution

= \frac{8}{25} arctan (t^{2} - 1) + C

\frac{d}{d x} (\frac{x ^{2} - 2 x - 3}{x + 2})

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y^{^{'}} - x yy = 2 x y

\int_{- 1}^{1} \frac{1}{( 1 + x ^{2} )} d x

What is the integral of 8/(5tsqrt(25t^2-25)) ?
The integral of 8/(5tsqrt(25t^2-25)) is 8/25 arctan(sqrt(t^2-1))+C