What is the integral of 1/(2x)+6/(x^3)-7sqrt(x) ?

The integral of 1/(2x)+6/(x^3)-7sqrt(x) is 1/2 ln|x|-3/(x^2)-14/3 x^{3/2}+C

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integral of 1/(2x)+6/(x^3)-7sqrt(x)

\int \frac{1}{2 x} + \frac{6}{x ^{3}} - 7 x d x

\frac{1}{2} ln ∣ x ∣ - \frac{3}{x ^{2}} - \frac{14}{3} x^{\frac{3}{2}} + C

Solution steps

\int \frac{1}{2 x} + \frac{6}{x ^{3}} - 7 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 \frac{1}{2 x} d x + \int \frac{6}{x ^{3}} d x - \int 7 x d x

\int \frac{1}{2 x} d x = \frac{1}{2} ln ∣ x ∣

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

\int 7 x d x = \frac{14}{3} x^{\frac{3}{2}}

= \frac{1}{2} ln ∣ x ∣ - \frac{3}{x ^{2}} - \frac{14}{3} x^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{2} ln ∣ x ∣ - \frac{3}{x ^{2}} - \frac{14}{3} x^{\frac{3}{2}} + C

\frac{\partial}{\partial x} (6 - 2 x)

\frac{d}{d x} ((x^{2})^{2})

\int sin^{7} (x) cos^{4} (x) d x

\int - B e^{- 2 t} d t

What is the integral of 1/(2x)+6/(x^3)-7sqrt(x) ?
The integral of 1/(2x)+6/(x^3)-7sqrt(x) is 1/2 ln|x|-3/(x^2)-14/3 x^{3/2}+C