To solve the integral
∫−221+2x1+x2dx,
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+2x1+x2 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+2x, which means du=2dx, or dx=2du.
When x=−2, u=1+2(−2)=−3.
When x=2, u=1+2(2)=5.
Therefore, the limits of integration change from [−2,2] to [−3,5].
Now, the integral becomes:
∫−35u1+(2u−1)2⋅21du.
Simplifying inside the integrand:
(2u−1)2=4(u−1)2=4u2−2u+1.
Thus,
1+(2u−1)2=1+4u2−2u+1=44+u2−2u+1=4u2−2u+5.
So, the integral becomes:
∫−35u4u2−2u+5⋅21du=81∫−35uu2−2u+5du=81∫−35(u−2+u5)du.
Breaking it into simpler integrals:
81(∫−35udu−2∫−35du+5∫−35u1du).
Solving each integral separately:
- ∫−35udu=[2u2]−35=225−29=216=8.
- −2∫−35du=−2[u]−35=−2(5−(−3))=−2(8)=−16.
- 5∫−35u1du=5[ln∣u∣]−35=5(ln5−ln3)=5ln35.
Combining these results:
81(8−16+5ln35)=81(−8+5ln35)=−1+85ln35.
Therefore, the value of the integral is:
∫−221+2x1+x2dx=−1+85ln35.
Would you like further details on any step or have any other questions? Here are 5 more questions you can ask:
- How do you determine if a function is even or odd?
- What are common techniques for evaluating definite integrals?
- Can you explain the process of u-substitution in integration?
- How do you handle improper integrals?
- 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.