What is the integral of 1/((x^2)*sqrt(25x^2+16)) ?

The integral of 1/((x^2)*sqrt(25x^2+16)) is -(sqrt(16+25x^2))/(16x)+C

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integral of 1/((x^2)*sqrt(25x^2+16))

\int \frac{1}{( x ^{2} ) \cdot 25 x ^{2} + 16} d x

- \frac{16 + 25 x ^{2}}{16 x} + C

Solution steps

\int \frac{1}{x ^{2} 25 x ^{2} + 16} d x

Apply Trigonometric Substitution

: \int \frac{5 sec ( u )}{16 tan ^{2} ( u )} d u

= \int \frac{5 sec ( u )}{16 tan ^{2} ( u )} d u

Take the constant out:

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

= \frac{5}{16} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{5}{16} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

= \frac{5}{16} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{5}{16} (- \frac{1}{v})

Substitute back

= \frac{5}{16} (- \frac{1}{sin ( arctan ( \frac{5}{4} x ) )})

Simplify

\frac{5}{16} (- \frac{1}{sin ( arctan ( \frac{5}{4} x ) )}) : - \frac{16 + 25 x ^{2}}{16 x}

= - \frac{16 + 25 x ^{2}}{16 x}

Add a constant to the solution

= - \frac{16 + 25 x ^{2}}{16 x} + C

\frac{\partial}{\partial y} (x y + 1)

\int \frac{1}{( x ^{3} - 8 ) ^{2}} d x

\int \frac{y ^{2} - 25}{y} d y

x \to 2 - lim (\frac{1}{x ^{3} - 2 x ^{2}})

\int x cos^{2} (2 x) d x

What is the integral of 1/((x^2)*sqrt(25x^2+16)) ?
The integral of 1/((x^2)*sqrt(25x^2+16)) is -(sqrt(16+25x^2))/(16x)+C