What is the integral of (sqrt(16+x^2))/(x^4) ?

The integral of (sqrt(16+x^2))/(x^4) is -(sqrt(16+x^2))/(3x^3)-(sqrt(16+x^2))/(48x)+C

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integral of (sqrt(16+x^2))/(x^4)

\int \frac{16 + x ^{2}}{x ^{4}} d x

- \frac{16 + x ^{2}}{3 x ^{3}} - \frac{16 + x ^{2}}{48 x} + C

Solution steps

\int \frac{16 + x ^{2}}{x ^{4}} d x

Apply Integration By Parts

: - \frac{16 + x ^{2}}{3 x ^{3}} - \int - \frac{1}{3 x ^{2} x ^{2} + 16} d x

= - \frac{16 + x ^{2}}{3 x ^{3}} - \int - \frac{1}{3 x ^{2} x ^{2} + 16} d x

\int - \frac{1}{3 x ^{2} x ^{2} + 16} d x = \frac{16 + x ^{2}}{48 x}

= - \frac{16 + x ^{2}}{3 x ^{3}} - \frac{16 + x ^{2}}{48 x}

Add a constant to the solution

= - \frac{16 + x ^{2}}{3 x ^{3}} - \frac{16 + x ^{2}}{48 x} + C

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What is the integral of (sqrt(16+x^2))/(x^4) ?
The integral of (sqrt(16+x^2))/(x^4) is -(sqrt(16+x^2))/(3x^3)-(sqrt(16+x^2))/(48x)+C