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

The integral of (sqrt(9+x^2))/(x^4) is -(sqrt(9+x^2))/(3x^3)-(sqrt(9+x^2))/(27x)+C

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

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

- \frac{9 + x ^{2}}{3 x ^{3}} - \frac{9 + x ^{2}}{27 x} + C

Solution steps

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

Apply Integration By Parts

: - \frac{9 + x ^{2}}{3 x ^{3}} - \int - \frac{1}{3 x ^{2} x ^{2} + 9} d x

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

\int - \frac{1}{3 x ^{2} x ^{2} + 9} d x = \frac{9 + x ^{2}}{27 x}

= - \frac{9 + x ^{2}}{3 x ^{3}} - \frac{9 + x ^{2}}{27 x}

Add a constant to the solution

= - \frac{9 + x ^{2}}{3 x ^{3}} - \frac{9 + x ^{2}}{27 x} + C

y = x^{3} - 9 x, (1, - 8)

\int_{0}^{1} (x^{2} + 1) e^{2 x} d x

6 x^{2} - 4 x y + 4 y^{2} - 14 = 0, (1, 2)

y = x + 1, (8, 3)

f (x) = x^{4} - 2 x^{3} + x - 1

What is the integral of (sqrt(9+x^2))/(x^4) ?
The integral of (sqrt(9+x^2))/(x^4) is -(sqrt(9+x^2))/(3x^3)-(sqrt(9+x^2))/(27x)+C