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

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

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

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

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

Solution steps

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

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} x)

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

Simplify

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

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

Add a constant to the solution

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

t^{2} (5 t + 2)^{3}

\int e^{4 x - 7} d x

f (x) = cot^{2} (x)

\int e^{0.04 x} d x

y^{^{'}} + y = 2 e^{- t}

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