What is the integral of (sqrt(36-x^2))/(x^4) ?

The integral of (sqrt(36-x^2))/(x^4) is -(sqrt(36-x^2))/(3x^3)+(sqrt(36-x^2))/(108x)+C

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integral of (sqrt(36-x^2))/(x^4)

\int \frac{36 - x ^{2}}{x ^{4}} d x

- \frac{36 - x ^{2}}{3 x ^{3}} + \frac{36 - x ^{2}}{108 x} + C

Solution steps

\int \frac{36 - x ^{2}}{x ^{4}} d x

Apply Integration By Parts

= - \frac{36 - x ^{2}}{3 x ^{3}} - \int \frac{1}{3 x ^{2} - x ^{2} + 36} d x

\int \frac{1}{3 x ^{2} - x ^{2} + 36} d x = - \frac{36 - x ^{2}}{108 x}

= - \frac{36 - x ^{2}}{3 x ^{3}} - (- \frac{36 - x ^{2}}{108 x})

Simplify

= - \frac{36 - x ^{2}}{3 x ^{3}} + \frac{36 - x ^{2}}{108 x}

Add a constant to the solution

= - \frac{36 - x ^{2}}{3 x ^{3}} + \frac{36 - x ^{2}}{108 x} + C

What is the integral of (sqrt(36-x^2))/(x^4) ?
The integral of (sqrt(36-x^2))/(x^4) is -(sqrt(36-x^2))/(3x^3)+(sqrt(36-x^2))/(108x)+C