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

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

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

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

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

Decimal

2.47763 \dots

Solution steps

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

Apply u-substitution

= \int_{2}^{1 + e^{2}} \frac{u - 1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{2}^{1 + e^{2}} \frac{u - 1}{u} d u

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= \frac{1}{2} \cdot \int_{2}^{1 + e^{2}} 1 - \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{2}^{1 + e^{2}} 1 d u - \int_{2}^{1 + e^{2}} \frac{1}{u} d u)

\int_{2}^{1 + e^{2}} 1 d u = e^{2} - 1

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

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

Simplify

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

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