What is the integral of 3x^3-4x^2+7/x-1/(x^2)+2e^x ?

The integral of 3x^3-4x^2+7/x-1/(x^2)+2e^x is (3x^4)/4-(4x^3)/3+7ln|x|+1/x+2e^x+C

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integral of 3x^3-4x^2+7/x-1/(x^2)+2e^x

\int 3 x^{3} - 4 x^{2} + \frac{7}{x} - \frac{1}{x ^{2}} + 2 e^{x} d x

\frac{3 x ^{4}}{4} - \frac{4 x ^{3}}{3} + 7 ln ∣ x ∣ + \frac{1}{x} + 2 e^{x} + C

Solution steps

\int 3 x^{3} - 4 x^{2} + \frac{7}{x} - \frac{1}{x ^{2}} + 2 e^{x} 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 3 x^{3} d x - \int 4 x^{2} d x + \int \frac{7}{x} d x - \int \frac{1}{x ^{2}} d x + \int 2 e^{x} d x

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

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int \frac{7}{x} d x = 7 ln ∣ x ∣

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

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

= \frac{3 x ^{4}}{4} - \frac{4 x ^{3}}{3} + 7 ln ∣ x ∣ - (- \frac{1}{x}) + 2 e^{x}

Simplify

= \frac{3 x ^{4}}{4} - \frac{4 x ^{3}}{3} + 7 ln ∣ x ∣ + \frac{1}{x} + 2 e^{x}

Add a constant to the solution

= \frac{3 x ^{4}}{4} - \frac{4 x ^{3}}{3} + 7 ln ∣ x ∣ + \frac{1}{x} + 2 e^{x} + C

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{3 x} (x + 1)

f (x) = 9 x - 8 x, (1, 1)

\int \frac{e ^{3 x}}{e ^{3 x} - e ^{- 3 x}} d x

f (x) = \frac{4}{x}, (- 4, - 1)

x \to - 6 lim (\frac{- 9 x}{x + 6})

What is the integral of 3x^3-4x^2+7/x-1/(x^2)+2e^x ?
The integral of 3x^3-4x^2+7/x-1/(x^2)+2e^x is (3x^4)/4-(4x^3)/3+7ln|x|+1/x+2e^x+C