What is the integral from 1 to e of (x^3+1)/(1+x^2) ?

The integral from 1 to e of (x^3+1)/(1+x^2) is 1/2 (e^2-1-ln(1+e^2)+ln(2))+arctan(e)-pi/4

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integral from 1 to e of (x^3+1)/(1+x^2)

\int_{1}^{e} \frac{x ^{3} + 1}{1 + x ^{2}} d x

\frac{1}{2} (e^{2} - 1 - ln (1 + e^{2}) + ln (2)) + arctan (e) - \frac{π}{4}

Decimal

2.91052 \dots

Solution steps

\int_{1}^{e} \frac{x ^{3} + 1}{1 + x ^{2}} d x

Expand

\frac{x ^{3} + 1}{1 + x ^{2}} : \frac{x ^{3}}{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_{1}^{e} \frac{x ^{3}}{1 + x ^{2}} d x + \int_{1}^{e} \frac{1}{1 + x ^{2}} d x

\int_{1}^{e} \frac{x ^{3}}{1 + x ^{2}} d x = \frac{1}{2} (e^{2} - 1 - ln (1 + e^{2}) + ln (2))

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

= \frac{1}{2} (e^{2} - 1 - ln (1 + e^{2}) + ln (2)) + arctan (e) - \frac{π}{4}

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

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

\int_{- 2}^{0} (x^{2} y - 2 x y) d y

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

\int_{- 4}^{3} ∣ x ∣ d x

What is the integral from 1 to e of (x^3+1)/(1+x^2) ?
The integral from 1 to e of (x^3+1)/(1+x^2) is 1/2 (e^2-1-ln(1+e^2)+ln(2))+arctan(e)-pi/4