Question:
$\int\left(\frac{x-a}{x}-\frac{x}{x+a}\right) dx$ is equal to
$\int\left(\frac{x-a}{x}-\frac{x}{x+a}\right) dx$ is equal to
Updated On: May 26, 2024
- $\log\left|\frac{x+a}{x}\right|+C$
\(a\log\left|\frac{x+a}{x}\right|+C\)
$a$ \(\log\left|\frac{x}{x+a}\right|+C\)
- $\log\left|\frac{x}{x+a}\right|+C$
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The Correct Option is B
Solution and Explanation
Let \(I=\int\left(\frac{x-a}{x}-\frac{x}{x+a}\right) d x\)
\(=\int \frac{\left(x^{2}-a^{2}\right)-x^{2}}{x(x+a)} d x\)
\(=-a^{2} \int \frac{1}{x(x+a)} d x\)
\(=\frac{-a^{2}}{a} \int\left[\frac{1}{x}-\frac{1}{x+a}\right] d x\)
\(=-a \log \left|\frac{x}{x+a}\right|+C\)
\(=a \log \left|\frac{x+a}{x}\right|+C\)
So, The correct option is B.
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Concepts Used:
Integrals of Some Particular Functions
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
Integrals of Some Particular Functions:
- ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
- ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
- ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
- ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
- ∫1/√(a2 – x2) dx = sin-1(x/a) + C
- ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C
These are tabulated below along with the meaning of each part.
