What is the integral of (x^5+2x^2-1)/(x^4) ?

The integral of (x^5+2x^2-1)/(x^4) is (x^2)/2-2/x+1/(3x^3)+C

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integral of (x^5+2x^2-1)/(x^4)

\int \frac{x ^{5} + 2 x ^{2} - 1}{x ^{4}} d x

\frac{x ^{2}}{2} - \frac{2}{x} + \frac{1}{3 x ^{3}} + C

Solution steps

\int \frac{x ^{5} + 2 x ^{2} - 1}{x ^{4}} d x

Expand

\frac{x ^{5} + 2 x ^{2} - 1}{x ^{4}} : x + \frac{2}{x ^{2}} - \frac{1}{x ^{4}}

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

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

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

\int \frac{1}{x ^{4}} d x = - \frac{1}{3 x ^{3}}

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

Simplify

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

Add a constant to the solution

= \frac{x ^{2}}{2} - \frac{2}{x} + \frac{1}{3 x ^{3}} + C

f (x) = (x - 2)^{\frac{1}{2}}, at x = 6

\int 5 sin^{3} (x) d x

t \to 1 lim (sin (- 3 π t))

y = e^{c o s (3) x^{2}}

x^{2} + x y + 2 y^{2} = 4, (1, 1)

What is the integral of (x^5+2x^2-1)/(x^4) ?
The integral of (x^5+2x^2-1)/(x^4) is (x^2)/2-2/x+1/(3x^3)+C