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

The integral of (sqrt(16+x^2))/(x^2) is -(sqrt(16+x^2))/x+ln(1/4 |x+sqrt(16+x^2)|)+C

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integral of (sqrt(16+x^2))/(x^2)

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

- \frac{16 + x ^{2}}{x} + ln (\frac{1}{4} x + 16 + x^{2}) + C

Solution steps

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

Apply Integration By Parts

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

\int - \frac{1}{16 + x ^{2}} d x = - ln (\frac{1}{4} x + 16 + x^{2})

= - \frac{16 + x ^{2}}{x} - (- ln (\frac{1}{4} x + 16 + x^{2}))

Simplify

= - \frac{16 + x ^{2}}{x} + ln (\frac{1}{4} x + 16 + x^{2})

Add a constant to the solution

= - \frac{16 + x ^{2}}{x} + ln (\frac{1}{4} x + 16 + x^{2}) + C

What is the integral of (sqrt(16+x^2))/(x^2) ?
The integral of (sqrt(16+x^2))/(x^2) is -(sqrt(16+x^2))/x+ln(1/4 |x+sqrt(16+x^2)|)+C