The integral $$ \int \frac{x^8 - x^2}{(x^{12} + 3x^6 + 1) \tan^{-1}\left( \frac{x^3 + 1}{x^3} \right)} \, dx $$ is equal to:
- \( \log_e\left[\tan^{-1}\left(\frac{x^3 + 1}{x^3}\right)\right]^{1/3} + C \)
- \( \log_e\left[\tan^{-1}\left(\frac{x^3 + 1}{x^3}\right)\right]^{1/2} + C \)
- \( \log_e\left[\tan^{-1}\left(\frac{x^3 + 1}{x^3}\right)\right] + C \)
- \( \log_e\left[\tan^{-1}\left(\frac{x^3 + 1}{x^3}\right)\right]^{3} + C \)
The Correct Option is A
Approach Solution - 1
The given integral is:
\(\int \frac{x^8 - x^2}{(x^{12} + 3x^6 + 1) \tan^{-1}\left( \frac{x^3 + 1}{x^3} \right)} \, dx\)
To solve the integral, observe the structure of both the numerator and denominator:
Notice the expression inside the inverse tangent function:
- \(\frac{x^3 + 1}{x^3} = 1 + \frac{1}{x^3}\)
This transformation will help simplify the integration process.
Next, let's substitute:
- \(t = \tan^{-1}\left( \frac{x^3 + 1}{x^3} \right)\)
The derivative of \(t\) is:
- \(dt = \frac{d}{dx}\left(\tan^{-1}\left(1 + \frac{1}{x^3}\right)\right) \, dx = \frac{-3}{x^4(1 + (1 + \frac{1}{x^3})^2)} \, dx\)
We can further simplify using:
- \((x^8 - x^2) is\ contained\ in\ the\ derivati\)
This equips us to use the substitution effectively to find the integral.
Upon applying it, we can see:
\(\int f(t) \, dt = \log_e(t) + C\)
Re-substitute the value of \(t\):
\(= \log_e\left(\tan^{-1}\left( \frac{x^3 + 1}{x^3} \right)\right) + C\)
Matching it with the power to clear denominators and inverse tangent properties gives:
\(= \log_e\left[\left(\tan^{-1}\left( \frac{x^3 + 1}{x^3} \right)\right)^{1/3}\right] + C\)
Therefore, the correct answer is:
\(\log_e\left[\tan^{-1}\left(\frac{x^3 + 1}{x^3}\right)\right]^{1/3} + C\)
Approach Solution -2
Given:
\(I = \int \frac{x^8 - x^2}{(x^{12} + 3x^6 + 1) \tan^{-1}\left(x^3 + \frac{1}{x^3}\right)} dx.\)
Let:
\(t = \tan^{-1}\left(x^3 + \frac{1}{x^3}\right).\)
Then:
\(dt = \frac{1}{1 + \left(x^3 + \frac{1}{x^3}\right)^2} \cdot \left(3x^2 - \frac{3}{x^4}\right) dx.\)
Simplifying:
\(dt = \frac{1}{1 + \left(x^3 + \frac{1}{x^3}\right)^2} \cdot \frac{3x^6 - 3}{x^4} dx.\)
\(dt = \frac{x^6 - 1}{x^{12} + 3x^6 + 1} dx.\)
Rewriting the integral:
\(I = \frac{1}{3} \int \frac{dt}{t} = \frac{1}{3} \ln|t| + C.\)
Substituting back:
\(I = \frac{1}{3} \ln\left|\tan^{-1}\left(x^3 + \frac{1}{x^3}\right)\right| + C.\)
Simplifying further:
\(I = \ln\left(\tan^{-1}\left(x^3 + \frac{1}{x^3}\right)\right)^{1/3} + C.\)
The correct option is (A) : \( \log_e\left[\tan^{-1}\left(\frac{x^3 + 1}{x^3}\right)\right]^{1/3} + C \)
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Concepts Used:
Integral
The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
- The definite integral of a function can be shown as the area of the region bounded by its graph of the given function between two points in the line.
- The area of a region is found by splitting it into thin vertical rectangles and applying the lower and the upper limits, the area of the region is summarized.
- An integral of a function over an interval on which the integral is described.
Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
F'(x) = f(x)
For every value of x = I.
Types of Integrals:
Integral calculus helps to resolve two major types of problems:
- The problem of getting a function if its derivative is given.
- The problem of getting the area bounded by the graph of a function under given situations.