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

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

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integral of (1+x^2)/x

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

ln ∣ x ∣ + \frac{x ^{2}}{2} + C

Solution steps

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

Expand

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ + \frac{x ^{2}}{2}

Add a constant to the solution

= ln ∣ x ∣ + \frac{x ^{2}}{2} + C

\int \frac{1}{5 x ( 6 + x )} d x

\frac{2 x + 2 x ^{2}}{( 1 + 2 x ) ^{2}}

\int 3 x sin^{2} (x^{2}) d x

- 4 cos (x)

12 y^{^{''}} - 12 y^{^{'}} + 6 y = 0

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