What is the integral of (sqrt(25x^2-3^2))/(4x^2) ?

The integral of (sqrt(25x^2-3^2))/(4x^2) is 1/4 (-(sqrt(25x^2-9))/x+5ln(1/3 |5x+sqrt(25x^2-9)|))+C

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integral of (sqrt(25x^2-3^2))/(4x^2)

\int \frac{25 x ^{2} - 3 ^{2}}{4 x ^{2}} d x

\frac{1}{4} (- \frac{25 x ^{2} - 9}{x} + 5 ln (\frac{1}{3} 5 x + 25 x^{2} - 9)) + C

Solution steps

\int \frac{25 x ^{2} - 3 ^{2}}{4 x ^{2}} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{25 x ^{2} - 3 ^{2}}{x ^{2}} d x

Simplify

= \frac{1}{4} \cdot \int \frac{25 x ^{2} - 9}{x ^{2}} d x

Apply Integration By Parts

: - \frac{25 x ^{2} - 9}{x} - \int - \frac{25}{25 x ^{2} - 9} d x

= \frac{1}{4} (- \frac{25 x ^{2} - 9}{x} - \int - \frac{25}{25 x ^{2} - 9} d x)

\int - \frac{25}{25 x ^{2} - 9} d x = - 5 ln (\frac{1}{3} 5 x + 25 x^{2} - 9)

= \frac{1}{4} (- \frac{25 x ^{2} - 9}{x} - (- 5 ln (\frac{1}{3} 5 x + 25 x^{2} - 9)))

Simplify

= \frac{1}{4} (- \frac{25 x ^{2} - 9}{x} + 5 ln (\frac{1}{3} 5 x + 25 x^{2} - 9))

Add a constant to the solution

= \frac{1}{4} (- \frac{25 x ^{2} - 9}{x} + 5 ln (\frac{1}{3} 5 x + 25 x^{2} - 9)) + C

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What is the integral of (sqrt(25x^2-3^2))/(4x^2) ?
The integral of (sqrt(25x^2-3^2))/(4x^2) is 1/4 (-(sqrt(25x^2-9))/x+5ln(1/3 |5x+sqrt(25x^2-9)|))+C