What is the integral of (sqrt(x^2+9))/(4x^2) ?

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

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integral of (sqrt(x^2+9))/(4x^2)

\int \frac{x ^{2} + 9}{4 x ^{2}} d x

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

Solution steps

\int \frac{x ^{2} + 9}{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{x ^{2} + 9}{x ^{2}} d x

Apply Integration By Parts

: - \frac{x ^{2} + 9}{x} - \int - \frac{1}{x ^{2} + 9} d x

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

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

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

Simplify

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

Add a constant to the solution

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

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