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

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

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

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

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

Solution steps

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

Simplify

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

Add a constant to the solution

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

\frac{8 t}{8 + t}

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\frac{\partial}{\partial x} (3 x^{\frac{1}{3}} y^{\frac{2}{3}})

x \to 0 lim (\frac{7}{1 + e ^{\frac{1}{x}}})

\int \frac{1}{7 x lo g _{3} ( x )} d x

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