What is the integral of 1/(r^2sqrt(4r^2-25)) ?

The integral of 1/(r^2sqrt(4r^2-25)) is (sqrt(4r^2-25))/(25r)+C

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integral of 1/(r^2sqrt(4r^2-25))

\int \frac{1}{r ^{2} 4 r ^{2} - 25} d r

\frac{4 r ^{2} - 25}{25 r} + C

Solution steps

\int \frac{1}{r ^{2} 4 r ^{2} - 25} d r

Apply Trigonometric Substitution

: \int \frac{2}{25 sec ( u )} d u

= \int \frac{2}{25 sec ( u )} d u

Take the constant out:

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

= \frac{2}{25} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{2}{25} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{2}{25} sin (u)

Substitute back

u = arcsec (\frac{2}{5} r)

= \frac{2}{25} sin (arcsec (\frac{2}{5} r))

Simplify

\frac{2}{25} sin (arcsec (\frac{2}{5} r)) : \frac{4 r ^{2} - 25}{25 r}

= \frac{4 r ^{2} - 25}{25 r}

Add a constant to the solution

= \frac{4 r ^{2} - 25}{25 r} + C

\int \frac{1}{sec ( x )} d x

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What is the integral of 1/(r^2sqrt(4r^2-25)) ?
The integral of 1/(r^2sqrt(4r^2-25)) is (sqrt(4r^2-25))/(25r)+C