What is the integral of (x^3+x+4)/(x^2) ?

The integral of (x^3+x+4)/(x^2) is (x^2)/2+ln|x|-4/x+C

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integral of (x^3+x+4)/(x^2)

\int \frac{x ^{3} + x + 4}{x ^{2}} d x

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

Solution steps

\int \frac{x ^{3} + x + 4}{x ^{2}} d x

Expand

\frac{x ^{3} + x + 4}{x ^{2}} : x + \frac{1}{x} + \frac{4}{x ^{2}}

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

Add a constant to the solution

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

\frac{d}{d y} (arctan (\frac{x}{y}))

\frac{d}{d x} (arctan (x^{4}))

f (t) = \frac{t ^{2} + 1}{t ^{2} - 1}

\int_{1}^{\infty} \frac{1}{5 x + 2} d x

What is the integral of (x^3+x+4)/(x^2) ?
The integral of (x^3+x+4)/(x^2) is (x^2)/2+ln|x|-4/x+C