What is the integral of (sqrt(49-x^2))/(x^4) ?

The integral of (sqrt(49-x^2))/(x^4) is -(sqrt(49-x^2))/(3x^3)+(sqrt(49-x^2))/(147x)+C

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integral of (sqrt(49-x^2))/(x^4)

\int \frac{49 - x ^{2}}{x ^{4}} d x

- \frac{49 - x ^{2}}{3 x ^{3}} + \frac{49 - x ^{2}}{147 x} + C

Solution steps

\int \frac{49 - x ^{2}}{x ^{4}} d x

Apply Integration By Parts

: - \frac{49 - x ^{2}}{3 x ^{3}} - \int \frac{1}{3 x ^{2} - x ^{2} + 49} d x

= - \frac{49 - x ^{2}}{3 x ^{3}} - \int \frac{1}{3 x ^{2} - x ^{2} + 49} d x

\int \frac{1}{3 x ^{2} - x ^{2} + 49} d x = - \frac{49 - x ^{2}}{147 x}

= - \frac{49 - x ^{2}}{3 x ^{3}} - (- \frac{49 - x ^{2}}{147 x})

Simplify

= - \frac{49 - x ^{2}}{3 x ^{3}} + \frac{49 - x ^{2}}{147 x}

Add a constant to the solution

= - \frac{49 - x ^{2}}{3 x ^{3}} + \frac{49 - x ^{2}}{147 x} + C

\frac{d ^{3}}{d x ^{3}} (x^{6} e^{x})

f (x) = \frac{6 x + 5}{x - 1}, at x = 2

t \to 2 lim (\frac{t - 2}{t ^{2} - 4})

\int 19 tan^{2} (x) sec^{3} (x) d x

\frac{\partial}{\partial x} (4 x^{3} + 2 x y + 1)

What is the integral of (sqrt(49-x^2))/(x^4) ?
The integral of (sqrt(49-x^2))/(x^4) is -(sqrt(49-x^2))/(3x^3)+(sqrt(49-x^2))/(147x)+C