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

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

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

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

ln (2) + \frac{π}{4}

Decimal

1.47854 \dots

Solution steps

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

Expand

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

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

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

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

= ln (2) + \frac{π}{4}

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