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

The integral of (-3/7 x^5+1/5 x^3-7/4 x^2) is -(x^6)/(14)+(x^4)/(20)-(7x^3)/(12)+C

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

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

- \frac{x ^{6}}{14} + \frac{x ^{4}}{20} - \frac{7 x ^{3}}{12} + C

Solution steps

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

\int \frac{3}{7} x^{5} d x = \frac{x ^{6}}{14}

\int \frac{1}{5} x^{3} d x = \frac{x ^{4}}{20}

\int \frac{7}{4} x^{2} d x = \frac{7 x ^{3}}{12}

= - \frac{x ^{6}}{14} + \frac{x ^{4}}{20} - \frac{7 x ^{3}}{12}

Add a constant to the solution

= - \frac{x ^{6}}{14} + \frac{x ^{4}}{20} - \frac{7 x ^{3}}{12} + C

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

\frac{d}{d x} (6 x + 3)

x \to 3 lim (\frac{x ^{2} + 1}{x + 3})

y^{^{'}} = y, y (1) = 1

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

What is the integral of (-3/7 x^5+1/5 x^3-7/4 x^2) ?
The integral of (-3/7 x^5+1/5 x^3-7/4 x^2) is -(x^6)/(14)+(x^4)/(20)-(7x^3)/(12)+C