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

The integral of (((x^3+1)^2)/(x^2)) is -((x^3+1)^2)/x+6((x^5)/5+(x^2)/2)+C

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

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

- \frac{( x ^{3} + 1 ) ^{2}}{x} + 6 (\frac{x ^{5}}{5} + \frac{x ^{2}}{2}) + C

Solution steps

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

Apply Integration By Parts

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

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

\int - 6 x (x^{3} + 1) d x = - 6 (\frac{x ^{5}}{5} + \frac{x ^{2}}{2})

= - \frac{( x ^{3} + 1 ) ^{2}}{x} - (- 6 (\frac{x ^{5}}{5} + \frac{x ^{2}}{2}))

Simplify

= - \frac{( x ^{3} + 1 ) ^{2}}{x} + 6 (\frac{x ^{5}}{5} + \frac{x ^{2}}{2})

Add a constant to the solution

= - \frac{( x ^{3} + 1 ) ^{2}}{x} + 6 (\frac{x ^{5}}{5} + \frac{x ^{2}}{2}) + C

\frac{\partial}{\partial x} (x y ln (x^{2} + y^{2}))

\frac{d}{d x} (ln (cosh (3 x)))

x \to \frac{π}{4} lim (\frac{sin ( 2 x ) - 1}{4 x - π})

\frac{d}{d x} (cos (2 x + 3 y))

y^{^{''}} - 16 y^{^{'}} + 63 y = 240 e^{x}

What is the integral of (((x^3+1)^2)/(x^2)) ?
The integral of (((x^3+1)^2)/(x^2)) is -((x^3+1)^2)/x+6((x^5)/5+(x^2)/2)+C