What is the integral of 7x^2-3x+8+1/x+2/(x^2) ?

The integral of 7x^2-3x+8+1/x+2/(x^2) is (7x^3)/3-(3x^2)/2+8x+ln|x|-2/x+C

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integral of 7x^2-3x+8+1/x+2/(x^2)

\int 7 x^{2} - 3 x + 8 + \frac{1}{x} + \frac{2}{x ^{2}} d x

\frac{7 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 8 x + ln ∣ x ∣ - \frac{2}{x} + C

Solution steps

\int 7 x^{2} - 3 x + 8 + \frac{1}{x} + \frac{2}{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^{2} d x - \int 3 x d x + \int 8 d x + \int \frac{1}{x} d x + \int \frac{2}{x ^{2}} d x

\int 7 x^{2} d x = \frac{7 x ^{3}}{3}

\int 3 x d x = \frac{3 x ^{2}}{2}

\int 8 d x = 8 x

\int \frac{1}{x} d x = ln ∣ x ∣

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

= \frac{7 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 8 x + ln ∣ x ∣ - \frac{2}{x}

Add a constant to the solution

= \frac{7 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 8 x + ln ∣ x ∣ - \frac{2}{x} + C

\frac{d}{d x} (e^{- x} - x^{2})

\int_{0}^{0.6} 400 x d x

in v erse l a pl a ce e^{- 5 s}

\frac{\partial}{\partial x} (\frac{y}{2 x + 3 y})

n = 1 \sum \infty \frac{( n + 4 )}{n ^{2}}

What is the integral of 7x^2-3x+8+1/x+2/(x^2) ?
The integral of 7x^2-3x+8+1/x+2/(x^2) is (7x^3)/3-(3x^2)/2+8x+ln|x|-2/x+C