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

The integral from 0 to 1 of (x+1)/(x^2+1) is 1/2 ln(2)+pi/4

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

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

\frac{1}{2} ln (2) + \frac{π}{4}

Decimal Notation

1.13197 \dots

Solution steps

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

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{0}^{1} \frac{x}{x ^{2} + 1} d x = \frac{1}{2} ln (2)

\int_{0}^{1} \frac{1}{x ^{2} + 1} d x = \frac{π}{4}

= \frac{1}{2} ln (2) + \frac{π}{4}

\int \frac{e ^{\frac{1}{x ^{8}}}}{x ^{9}} d x

x \to - 3 lim (\frac{x - 3}{9 x})

\frac{\partial}{\partial x} (x^{2} + 5 x z + 3 y z^{2})

f (x) = 8 ln (t^{8})

y = \frac{cos ( x )}{x ^{5}}

What is the integral from 0 to 1 of (x+1)/(x^2+1) ?
The integral from 0 to 1 of (x+1)/(x^2+1) is 1/2 ln(2)+pi/4