What is the integral of (7x^3-1)/(x^2) ?

The integral of (7x^3-1)/(x^2) is (7x^2)/2+1/x+C

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integral of (7x^3-1)/(x^2)

\int \frac{7 x ^{3} - 1}{x ^{2}} d x

\frac{7 x ^{2}}{2} + \frac{1}{x} + C

Solution steps

\int \frac{7 x ^{3} - 1}{x ^{2}} d x

Expand

\frac{7 x ^{3} - 1}{x ^{2}} : 7 x - \frac{1}{x ^{2}}

= \int 7 x - \frac{1}{x ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 7 x d x - \int \frac{1}{x ^{2}} d x

\int 7 x d x = \frac{7 x ^{2}}{2}

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

= \frac{7 x ^{2}}{2} - (- \frac{1}{x})

Simplify

= \frac{7 x ^{2}}{2} + \frac{1}{x}

Add a constant to the solution

= \frac{7 x ^{2}}{2} + \frac{1}{x} + C

\frac{\partial}{\partial x} (e^{x} sec^{2} (y))

f (x) = \frac{2 x}{x ^{2} + 9}

\int I n^{2} x d x

f (x) = 4 e^{5 x^{2}} (5 - x)^{7}

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

What is the integral of (7x^3-1)/(x^2) ?
The integral of (7x^3-1)/(x^2) is (7x^2)/2+1/x+C