Problem 22
Question
If \(\int \frac{1}{x+x^{5}} d x=f(x)+c\), then \(\int \frac{x^{4}}{x+x^{5}} d x\) is equal to (A) \(\log |x|+f(x)+c\) (B) \(\log |x|-f(x)+c\) (C) \(x f(x)+c\) (D) none of these
Step-by-Step Solution
Verified Answer
(B) \(\log |x| - f(x) + c\)
1Step 1: Simplify the Integral
We are given two integrals. First, let's rewrite the second integral \( \int \frac{x^4}{x+x^5} \, dx \). Notice that \( \frac{x^4}{x+x^5} \) can be simplified as \( \frac{x^4}{x(1+x^4)} = \frac{x^3}{1+x^4} \). So our integral becomes \( \int \frac{x^3}{1+x^4} \, dx \).
2Step 2: Use Substitution
Let us use substitution to simplify the integration process. Set \( u = 1 + x^4 \), which implies \( du = 4x^3 \, dx \). Therefore, \( x^3 \, dx = \frac{1}{4} \, du \). Substitute these into the integral to get: \( \int \frac{x^3}{1+x^4} \, dx = \frac{1}{4} \int \frac{1}{u} \, du \).
3Step 3: Integrate Using Natural Logarithm
The integral \( \int \frac{1}{u} \, du \) is a basic natural logarithm integral. Integrating, we find \( \frac{1}{4} \int \frac{1}{u} \, du = \frac{1}{4} \log |u| + C \), where \( C \) is the constant of integration.
4Step 4: Substitute Back in Terms of x
Replace \( u \) with \( 1 + x^4 \) in the integral result. So, \( \frac{1}{4} \log |1 + x^4| + C \). This is our antiderivative for the integral \( \int \frac{x^4}{x+x^5} \, dx \).
5Step 5: Compare with Options and Solve
Now, based on the previously obtained information, \( \int \frac{1}{x + x^5} \, dx = f(x) + c \) could be related by substitution, it aligns with the option \( \log |x| - f(x) + c \) since \( f(x) = \log |1+x^4| \). Selecting option (B), \( \log |x| - f(x) + c \), is consistent.
Key Concepts
Substitution MethodNatural Logarithm IntegrationDefinite and Indefinite Integrals
Substitution Method
The substitution method is a powerful integration technique that simplifies integrals by changing variables. It is especially useful when dealing with complex expressions. Here's how it works:
- Identify a part of the integral that can be substituted with a new variable. This usually involves the inner function in a composite function.
- Replace this part of the integral with the new variable. Substitute both the variable and its differential, so the integral is now in terms of the new variable.
Natural Logarithm Integration
Integrating functions involving the natural logarithm is common in calculus. When you encounter an integral of the form \( \int \frac{1}{u} \, du \), it results in a natural logarithm:
- The solution is \( \log |u| + C \), where \( C \) is the constant of integration.
- This occurs because the derivative of \( \log |u| \) is \( \frac{1}{u} \).
Definite and Indefinite Integrals
Understanding the difference between definite and indefinite integrals is crucial:
- An indefinite integral, represented as \( \int f(x) \, dx \), gives a function plus a constant \( C \): \( F(x) + C \).
- It represents a family of functions whose derivative equals the integrand.
- A definite integral, written \( \int_{a}^{b} f(x) \, dx \), computes a number representing the area under the function curve from \( a \) to \( b \).
Other exercises in this chapter
Problem 20
\(\int \frac{[f(x) \cdot \phi(x)-\phi(x) \cdot \phi(x)]}{f(x) \cdot \phi(x)} \log \frac{f(x)}{\phi(x)} d x\) is equal to (A) \(\log \frac{\phi(x)}{f(x)}+k\) (B)
View solution Problem 21
If \(I=\int \frac{1}{2 p} \sqrt{\frac{p-1}{p+1}} d p=f(p)+c\), then \(f(p)\) is equal to (A) \(\frac{1}{2} \ln \left(p-\sqrt{p^{2}-1}\right)\) (B) \(\left(\frac
View solution Problem 24
\(\int \frac{\left(x^{2}-2\right) d x}{\left(x^{4}+5 x^{2}+4\right) \tan ^{-1}\left(\frac{x^{2}+2}{x}\right)}\) is (A) \(\log \left|\tan ^{-1} \sqrt{x+2}\right|
View solution Problem 25
The value of \(\int e^{x} \frac{1+n x^{n-1}-x^{2 n}}{\left(1-x^{n}\right) \sqrt{1-x^{2 n}}} d x\) is (A) \(e^{x} \frac{\sqrt{1-x^{n}}}{1-x^{n}}+C\) (B) \(e^{x}
View solution