Math Problem Statement

solve

Solution

To solve the integral

221+x21+2xdx,\int_{-2}^{2} \frac{1 + x^2}{1 + 2x} \, dx,

we can use symmetry properties and substitution if necessary. Let's examine the function inside the integral for possible simplifications.

First, note that the integrand 1+x21+2x\frac{1 + x^2}{1 + 2x} is not an even or odd function, which means we can't directly simplify it using symmetry around the y-axis.

However, we can consider a substitution to simplify the integration process. Let's try u=1+2xu = 1 + 2x, which means du=2dxdu = 2dx, or dx=du2dx = \frac{du}{2}.

When x=2x = -2, u=1+2(2)=3u = 1 + 2(-2) = -3.

When x=2x = 2, u=1+2(2)=5u = 1 + 2(2) = 5.

Therefore, the limits of integration change from [2,2][-2, 2] to [3,5][-3, 5].

Now, the integral becomes:

351+(u12)2u12du.\int_{-3}^{5} \frac{1 + \left( \frac{u-1}{2} \right)^2}{u} \cdot \frac{1}{2} \, du.

Simplifying inside the integrand:

(u12)2=(u1)24=u22u+14.\left( \frac{u-1}{2} \right)^2 = \frac{(u-1)^2}{4} = \frac{u^2 - 2u + 1}{4}.

Thus,

1+(u12)2=1+u22u+14=4+u22u+14=u22u+54.1 + \left( \frac{u-1}{2} \right)^2 = 1 + \frac{u^2 - 2u + 1}{4} = \frac{4 + u^2 - 2u + 1}{4} = \frac{u^2 - 2u + 5}{4}.

So, the integral becomes:

35u22u+54u12du=1835u22u+5udu=1835(u2+5u)du.\int_{-3}^{5} \frac{\frac{u^2 - 2u + 5}{4}}{u} \cdot \frac{1}{2} \, du = \frac{1}{8} \int_{-3}^{5} \frac{u^2 - 2u + 5}{u} \, du = \frac{1}{8} \int_{-3}^{5} \left( u - 2 + \frac{5}{u} \right) \, du.

Breaking it into simpler integrals:

18(35udu235du+5351udu).\frac{1}{8} \left( \int_{-3}^{5} u \, du - 2 \int_{-3}^{5} du + 5 \int_{-3}^{5} \frac{1}{u} \, du \right).

Solving each integral separately:

  1. 35udu=[u22]35=25292=162=8.\int_{-3}^{5} u \, du = \left[ \frac{u^2}{2} \right]_{-3}^{5} = \frac{25}{2} - \frac{9}{2} = \frac{16}{2} = 8.
  2. 235du=2[u]35=2(5(3))=2(8)=16.-2 \int_{-3}^{5} du = -2[u]_{-3}^{5} = -2(5 - (-3)) = -2(8) = -16.
  3. 5351udu=5[lnu]35=5(ln5ln3)=5ln53.5 \int_{-3}^{5} \frac{1}{u} \, du = 5[\ln|u|]_{-3}^{5} = 5(\ln 5 - \ln 3) = 5 \ln \frac{5}{3}.

Combining these results:

18(816+5ln53)=18(8+5ln53)=1+58ln53.\frac{1}{8} \left( 8 - 16 + 5 \ln \frac{5}{3} \right) = \frac{1}{8} \left( -8 + 5 \ln \frac{5}{3} \right) = -1 + \frac{5}{8} \ln \frac{5}{3}.

Therefore, the value of the integral is:

221+x21+2xdx=1+58ln53.\int_{-2}^{2} \frac{1 + x^2}{1 + 2x} \, dx = -1 + \frac{5}{8} \ln \frac{5}{3}.

Would you like further details on any step or have any other questions? Here are 5 more questions you can ask:

  1. How do you determine if a function is even or odd?
  2. What are common techniques for evaluating definite integrals?
  3. Can you explain the process of u-substitution in integration?
  4. How do you handle improper integrals?
  5. What are some applications of definite integrals in real life?

Tip: Always double-check the limits of integration when performing a substitution to ensure accuracy.

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Math Problem Analysis

Mathematical Concepts

Integration
Substitution
Definite Integrals

Formulas

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Theorems

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Suitable Grade Level

Advanced College Level