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

The integral of (sqrt(4-x^2))/(x^4) is -(sqrt(4-x^2))/(3x^3)+(sqrt(4-x^2))/(12x)+C

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

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

- \frac{4 - x ^{2}}{3 x ^{3}} + \frac{4 - x ^{2}}{12 x} + C

Solution steps

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

Apply Integration By Parts

: - \frac{4 - x ^{2}}{3 x ^{3}} - \int \frac{1}{3 x ^{2} - x ^{2} + 4} d x

= - \frac{4 - x ^{2}}{3 x ^{3}} - \int \frac{1}{3 x ^{2} - x ^{2} + 4} d x

\int \frac{1}{3 x ^{2} - x ^{2} + 4} d x = - \frac{4 - x ^{2}}{12 x}

= - \frac{4 - x ^{2}}{3 x ^{3}} - (- \frac{4 - x ^{2}}{12 x})

Simplify

= - \frac{4 - x ^{2}}{3 x ^{3}} + \frac{4 - x ^{2}}{12 x}

Add a constant to the solution

= - \frac{4 - x ^{2}}{3 x ^{3}} + \frac{4 - x ^{2}}{12 x} + C

ln (ln (8 x))

\frac{- x ^{3}}{y ^{3}}

y = 4 x, y = 2 x^{2}

14 x^{3} + 7 x^{2} + 11 x - 29

\frac{d y}{d x} = \frac{4 x sec ( \frac{y}{x} ) + y}{x}

What is the integral of (sqrt(4-x^2))/(x^4) ?
The integral of (sqrt(4-x^2))/(x^4) is -(sqrt(4-x^2))/(3x^3)+(sqrt(4-x^2))/(12x)+C