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

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

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

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

4 ln ∣ x ∣ - \frac{1}{x} + C

Solution steps

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

Expand

\frac{4 x + 1}{x ^{2}} : \frac{4}{x} + \frac{1}{x ^{2}}

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

\int \frac{4}{x} d x = 4 ln ∣ x ∣

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

= 4 ln ∣ x ∣ - \frac{1}{x}

Add a constant to the solution

= 4 ln ∣ x ∣ - \frac{1}{x} + C

y^{^{'}} = 3 + t - y, y (0) = 1

\int_{1}^{3} (x^{2} - 5 x + 3) d x

- 1

\frac{d}{d x} (cos^{2} (e^{c o s^{2} (x)}))

\int ln (cos (x)) sin (x) d x

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