What is the integral of-3/(5tsqrt(25t^2-49)) ?

The integral of-3/(5tsqrt(25t^2-49)) is -3/35 arcsec(5/7 t)+C

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integral of-3/(5tsqrt(25t^2-49))

\int - \frac{3}{5 t 25 t ^{2} - 49} d t

- \frac{3}{35} arcsec (\frac{5}{7} t) + C

Solution steps

\int - \frac{3}{5 t 25 t ^{2} - 49} d t

Take the constant out:

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

= - \frac{3}{5} \cdot \int \frac{1}{t 25 t ^{2} - 49} d t

Apply Trigonometric Substitution

: \int \frac{1}{7} d u

= - \frac{3}{5} \cdot \int \frac{1}{7} d u

Take the constant out:

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

= - \frac{3}{5} \cdot \frac{1}{7} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= - \frac{3}{5} \cdot \frac{1}{7} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{5}{7} t)

= - \frac{3}{5} \cdot \frac{1}{7} \cdot 1 \cdot arcsec (\frac{5}{7} t)

Simplify

- \frac{3}{5} \cdot \frac{1}{7} \cdot 1 \cdot arcsec (\frac{5}{7} t) : - \frac{3}{35} arcsec (\frac{5}{7} t)

= - \frac{3}{35} arcsec (\frac{5}{7} t)

Add a constant to the solution

= - \frac{3}{35} arcsec (\frac{5}{7} t) + C

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What is the integral of-3/(5tsqrt(25t^2-49)) ?
The integral of-3/(5tsqrt(25t^2-49)) is -3/35 arcsec(5/7 t)+C