Problem 28
Question
Find the integral. $$ \int \frac{x^{3}+x+1}{x^{4}+2 x^{2}+1} d x $$
Step-by-Step Solution
Verified Answer
The integral of \(\int \frac{x^{3}+x+1}{x^{4}+2 x^{2}+1} dx\) is \(\frac{1}{2} [\ln|x^{2}+1 |-x^{2} - 1]\).
1Step 1: Recognize the denominator as a perfect square
Rewrite the denominator as a perfect square. In this case, \(x^{4}+2x^{2}+1\) can be written as \((x^{2}+1)^{2}\).
2Step 2: Substitute
Let \(u = x^{2} + 1\). Then, \(du = 2x dx\) and \(x dx = \frac{1}{2} du\). Replace the variable x and dx with u and du in the integral equation, respectively.
3Step 3: Rewrite the integral
Substitution should yield the following equation: \(\int \frac{u-1}{u^{2}}* \frac{1}{2} du\).
4Step 4: Simplify the integral
Now divide the integrand into two parts and then integrate each part separately. This gives us: \(\frac{1}{2} \int(\frac{1}{u}-\frac{1}{u^{2}}) du\).
5Step 5: Solve the integral
Solving both integrals will give us: \(\frac{1}{2}[\ln|u| - u]\).
6Step 6: Substitute u back
Substitute 'u' back with 'x': The final answer is \(\frac{1}{2}[\ln|x^{2} +1 |- x^{2} - 1]\).
Other exercises in this chapter
Problem 28
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In Exercises \(27-38,\) (a) describe the type of indeterminate form (if any) that is obtained by direct substitution. (b) Evaluate the limit, using L'Hôpital's
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