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

The integral from 0 to 1 of (x+1)/(4+x^2) is -ln(2)+(ln(5)+arctan(1/2))/2

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integral from 0 to 1 of (x+1)/(4+x^2)

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

- ln (2) + \frac{ln ( 5 ) + arctan ( \frac{1}{2} )}{2}

Decimal Notation

0.34339 \dots

Solution steps

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

Expand

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

\int_{0}^{1} \frac{x}{4 + x ^{2}} d x = \frac{1}{2} ln (5) - ln (2)

\int_{0}^{1} \frac{1}{4 + x ^{2}} d x = \frac{1}{2} arctan (\frac{1}{2})

= \frac{1}{2} ln (5) - ln (2) + \frac{1}{2} arctan (\frac{1}{2})

Simplify

\frac{1}{2} ln (5) - ln (2) + \frac{1}{2} arctan (\frac{1}{2}) : - ln (2) + \frac{ln ( 5 ) + arctan ( \frac{1}{2} )}{2}

= - ln (2) + \frac{ln ( 5 ) + arctan ( \frac{1}{2} )}{2}

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What is the integral from 0 to 1 of (x+1)/(4+x^2) ?
The integral from 0 to 1 of (x+1)/(4+x^2) is -ln(2)+(ln(5)+arctan(1/2))/2