What is the integral of (-5x^3-1/x+5/(x^4)+6sqrt(x)) ?

The integral of (-5x^3-1/x+5/(x^4)+6sqrt(x)) is -(5x^4)/4-ln|x|-5/(3x^3)+4x^{3/2}+C

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integral of (-5x^3-1/x+5/(x^4)+6sqrt(x))

\int (- 5 x^{3} - \frac{1}{x} + \frac{5}{x ^{4}} + 6 x) d x

- \frac{5 x ^{4}}{4} - ln ∣ x ∣ - \frac{5}{3 x ^{3}} + 4 x^{\frac{3}{2}} + C

Solution steps

\int - 5 x^{3} - \frac{1}{x} + \frac{5}{x ^{4}} + 6 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 5 x^{3} d x - \int \frac{1}{x} d x + \int \frac{5}{x ^{4}} d x + \int 6 x d x

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

\int 6 x d x = 4 x^{\frac{3}{2}}

= - \frac{5 x ^{4}}{4} - ln ∣ x ∣ - \frac{5}{3 x ^{3}} + 4 x^{\frac{3}{2}}

Add a constant to the solution

= - \frac{5 x ^{4}}{4} - ln ∣ x ∣ - \frac{5}{3 x ^{3}} + 4 x^{\frac{3}{2}} + C

\int x + 6 d x

\int t ln (t) d t

\frac{\partial}{\partial x} (ln (x^{2} - y^{2}))

\int_{1}^{2} \frac{2 x + 1}{x ^{2} + x} d x

\int_{1}^{3} 4 x^{3} - 8 x + 1 d x

What is the integral of (-5x^3-1/x+5/(x^4)+6sqrt(x)) ?
The integral of (-5x^3-1/x+5/(x^4)+6sqrt(x)) is -(5x^4)/4-ln|x|-5/(3x^3)+4x^{3/2}+C