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

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

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integral of x/(1+x^2)

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

\frac{1}{2} ln 1 + x^{2} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u} d u

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 + x^{2}

= \frac{1}{2} ln 1 + x^{2}

Add a constant to the solution

= \frac{1}{2} ln 1 + x^{2} + C

y^{^{''}} - 4 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = - 8

f (x) = \frac{x ^{3}}{9}

\frac{d y}{d x} - 3 y = - 7 sin (x) - 7 cos (x)

x \to + 1 lim (e^{x - 1} - 1)

cos (x) + x sin (x)

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