What is the integral from 0 to R of x/(1+x^2) ?

The integral from 0 to R of x/(1+x^2) is 1/2 ln|R^2+1|

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integral from 0 to R of x/(1+x^2)

\int_{0}^{R} \frac{x}{1 + x ^{2}} d x

\frac{1}{2} ln R^{2} + 1

Solution steps

\int_{0}^{R} \frac{x}{1 + x ^{2}} d x

Apply u-substitution

: \int_{1}^{1 + R^{2}} \frac{1}{2 u} d u

= \int_{1}^{1 + R^{2}} \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_{1}^{1 + R^{2}} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{1}^{1 + R^{2}}

Compute the boundaries:

ln R^{2} + 1

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

2 y^{^{''}} - 5 y^{^{'}} - 3 y = 0

\int \frac{i n ^{2} x}{x} d x

\frac{\partial}{\partial x} (3 x^{4} y^{2} + 1)

x \to 2 lim (\frac{2}{( x - 2 ) ^{2}})

\frac{\partial}{\partial y} (x^{5} + x^{2} y + x + 1)

What is the integral from 0 to R of x/(1+x^2) ?
The integral from 0 to R of x/(1+x^2) is 1/2 ln|R^2+1|