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

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

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

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

\frac{π}{4} + \frac{1}{2} ln (2)

Decimal

1.13197 \dots

Solution steps

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

Expand

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

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

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

= \frac{π}{4} + \frac{1}{2} ln (2)

y = x^{5} sin^{2} (2 x)

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

x y^{^{'}} - y = 2 x ln (x)

f (x) = \frac{x ^{2}}{x + 3}, (3, \frac{3}{2})

\int tan^{2} (x) sec (x) d x

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