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

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

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

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

ln (2) + 1

Decimal Notation

1.69314 \dots

Solution steps

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

\frac{( x + 1 ) ^{2}}{x ^{2} + 1} = \frac{2 x}{x ^{2} + 1} + 1

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

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

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

\int_{0}^{1} 1 d x = 1

= ln (2) + 1

sin (e^{2 x})

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\frac{d}{d a} ((3 a^{2})^{3})

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