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

The integral of 1/(sqrt(x))+4/(x^2)-1 is 2sqrt(x)-4/x-x+C

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integral of 1/(sqrt(x))+4/(x^2)-1

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

2 x - \frac{4}{x} - x + C

Solution steps

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

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

\int \frac{4}{x ^{2}} d x = - \frac{4}{x}

\int 1 d x = x

= 2 x - \frac{4}{x} - x

Add a constant to the solution

= 2 x - \frac{4}{x} - x + C

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