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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

2 x y - 9 x^{2} + (2 y + x^{2} + 1) \frac{d y}{d x} = 0, y (0) = 3

\int 6 t^{5} d t

x^{2} y^{^{'}} - 3 x y - 2 y^{2} = 0

\frac{d y}{d t} + \frac{2}{t} y = t - 1 + \frac{y}{t}

y = 9 x^{\frac{5}{2}} - 7 x^{\frac{3}{2}}, at x = 4

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