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

The integral of 5x(x^4-1/5) is 5(1/10 x^2(5x^4-1)-(x^6)/3)+C

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

\int 5 x (x^{4} - \frac{1}{5}) d x

5 (\frac{1}{10} x^{2} (5 x^{4} - 1) - \frac{x ^{6}}{3}) + C

Solution steps

\int 5 x (x^{4} - \frac{1}{5}) d x

Take the constant out:

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

= 5 \cdot \int x (x^{4} - \frac{1}{5}) d x

Apply Integration By Parts

: \frac{1}{10} x^{2} (5 x^{4} - 1) - \int 2 x^{5} d x

= 5 (\frac{1}{10} x^{2} (5 x^{4} - 1) - \int 2 x^{5} d x)

\int 2 x^{5} d x = \frac{x ^{6}}{3}

= 5 (\frac{1}{10} x^{2} (5 x^{4} - 1) - \frac{x ^{6}}{3})

Add a constant to the solution

= 5 (\frac{1}{10} x^{2} (5 x^{4} - 1) - \frac{x ^{6}}{3}) + C

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x \to \infty lim ((4^{x} + 4^{x})^{\frac{1}{2 x}})

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What is the integral of 5x(x^4-1/5) ?
The integral of 5x(x^4-1/5) is 5(1/10 x^2(5x^4-1)-(x^6)/3)+C