What is the integral of (2x^4-3x^3+5)/(7x^2) ?

The integral of (2x^4-3x^3+5)/(7x^2) is 1/7 ((2x^3)/3-(3x^2)/2-5/x)+C

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integral of (2x^4-3x^3+5)/(7x^2)

\int \frac{2 x ^{4} - 3 x ^{3} + 5}{7 x ^{2}} d x

\frac{1}{7} (\frac{2 x ^{3}}{3} - \frac{3 x ^{2}}{2} - \frac{5}{x}) + C

Solution steps

\int \frac{2 x ^{4} - 3 x ^{3} + 5}{7 x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{7} \cdot \int \frac{2 x ^{4} - 3 x ^{3} + 5}{x ^{2}} d x

Expand

\frac{2 x ^{4} - 3 x ^{3} + 5}{x ^{2}} : 2 x^{2} - 3 x + \frac{5}{x ^{2}}

= \frac{1}{7} \cdot \int 2 x^{2} - 3 x + \frac{5}{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

= \frac{1}{7} (\int 2 x^{2} d x - \int 3 x d x + \int \frac{5}{x ^{2}} d x)

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

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

\int \frac{5}{x ^{2}} d x = - \frac{5}{x}

= \frac{1}{7} (\frac{2 x ^{3}}{3} - \frac{3 x ^{2}}{2} - \frac{5}{x})

Add a constant to the solution

= \frac{1}{7} (\frac{2 x ^{3}}{3} - \frac{3 x ^{2}}{2} - \frac{5}{x}) + C

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What is the integral of (2x^4-3x^3+5)/(7x^2) ?
The integral of (2x^4-3x^3+5)/(7x^2) is 1/7 ((2x^3)/3-(3x^2)/2-5/x)+C