Evaluate the following definite integral. ∫01x3+1x+1dx

Question

Evaluate the following definite integral.  ∫01x3+1x+1dx

John Twitchell · Accepted Answer

Step 1
Definite integrals can be computed using the fundamental theorem of calculus. It states that if

F (x) = \int f (x) d x

then we have

\int_{a}^{b} f (x) d x = F (b) - F (a)

.
We will use the following identity for this problem:

x^{3} + 1 = (x + 1) (x^{2} - x + 1)

. Use this to simplify the integrand. Use the standard integral

\int x^{n} d x = \frac{x^{n + 1}}{n + 1} f or n \neq - 1

.
Step 2
The integral to be computed is

\int_{0}^{1} \frac{x^{3} + 1}{x + 1} d x

. Use information from step 1 to simplify and integrate.

\int_{0}^{1} \frac{x^{3} + 1}{x + 1} d x = \int_{0}^{1} \frac{(x + 1) (x^{2} - x + 1)}{x + 1} d x

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

= {(\frac{x^{3}}{3} - \frac{x^{2}}{2} + x)}_{0}^{1}

= \frac{1}{3} - \frac{1}{2} + 1

= \frac{2 - 3 + 6}{6}

= \frac{5}{6}

Hence, the definite integral is equal to

\frac{5}{6}

.

Evaluate the following definite integral. \int_{0}^{1}\frac{x^{3}+1}{x+1}dx