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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

Add a constant to the solution

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

x \to 1 lim (x^{2} + 3 x - 4)

\frac{( x ^{2} e ^{x} )}{x ^{2} + e ^{x}}

\frac{d}{d x} (5 e^{x} (cos (x) - sin (x)))

\frac{d y}{d x} = (x - 8) e^{- 2 y}, y (8) = ln (8)

\frac{d}{d x} (tan^{4} (x))

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