What is the integral of ((x^4+1)^2)/(x^3) ?

The integral of ((x^4+1)^2)/(x^3) is -((x^4+1)^2}{2x^2}+4(1/2 x^2(x^4+1)-\frac{x^6)/3)+C

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integral of ((x^4+1)^2)/(x^3)

\int \frac{( x ^{4} + 1 ) ^{2}}{x ^{3}} d x

- \frac{( x ^{4} + 1 ) ^{2}}{2 x ^{2}} + 4 (\frac{1}{2} x^{2} (x^{4} + 1) - \frac{x ^{6}}{3}) + C

Solution steps

\int \frac{( x ^{4} + 1 ) ^{2}}{x ^{3}} d x

Apply Integration By Parts

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

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

\int - 4 x (x^{4} + 1) d x = - 4 (\frac{1}{2} x^{2} (x^{4} + 1) - \frac{x ^{6}}{3})

= - \frac{( x ^{4} + 1 ) ^{2}}{2 x ^{2}} - (- 4 (\frac{1}{2} x^{2} (x^{4} + 1) - \frac{x ^{6}}{3}))

Simplify

= - \frac{( x ^{4} + 1 ) ^{2}}{2 x ^{2}} + 4 (\frac{1}{2} x^{2} (x^{4} + 1) - \frac{x ^{6}}{3})

Add a constant to the solution

= - \frac{( x ^{4} + 1 ) ^{2}}{2 x ^{2}} + 4 (\frac{1}{2} x^{2} (x^{4} + 1) - \frac{x ^{6}}{3}) + C

y = \frac{5}{x}

\frac{\partial}{\partial x} (y d x)

\int x ln (\frac{x + 2}{x + 1}) d x

\int \frac{4 x}{y} d x

\frac{d}{d x} (x \cdot e^{- x})

What is the integral of ((x^4+1)^2)/(x^3) ?
The integral of ((x^4+1)^2)/(x^3) is -((x^4+1)^2}{2x^2}+4(1/2 x^2(x^4+1)-\frac{x^6)/3)+C