$\int_{0}^{1} \frac{1}{\left(x^{2}+16\right)\left(x^{2}+25\right)} \,dx$ is equal to
- $\frac{1}{5} \left[\frac{1}{4} tan^{-1}\left(\frac{1}{4}\right)-\frac{1}{5} tan^{-1}\left(\frac{1}{5}\right)\right]$
- $\frac{1}{9} \left[\frac{1}{4} tan^{-1}\left(\frac{1}{4}\right)-\frac{1}{5} tan^{-1}\left(\frac{1}{5}\right)\right]$
- $\frac{1}{4} \left[\frac{1}{4} tan^{-1}\left(\frac{1}{4}\right)-\frac{1}{5} tan^{-1}\left(\frac{1}{5}\right)\right]$
- $\frac{1}{9} \left[\frac{1}{5} tan^{-1}\left(\frac{1}{4}\right)-\frac{1}{4} tan^{-1}\left(\frac{1}{5}\right)\right]$
The Correct Option is B
Solution and Explanation
The correct option is (B): \(\frac{1}{9} \left[\frac{1}{4} tan^{-1}\left(\frac{1}{4}\right)-\frac{1}{5} tan^{-1}\left(\frac{1}{5}\right)\right]\).
\(I =\int_{0}^{1} \frac{1}{\left(x^{2}+16\right)\left(x^{2}+25\right)} d x\)
\(=\frac{1}{9} \int_{0}^{1}\left(\frac{1}{x^{2}+16}-\frac{1}{x^{2}+25}\right) d x\)
\(=\frac{1}{9}\left[\frac{1}{4} \tan ^{-1} \frac{x}{4}-\frac{1}{5} \tan ^{-1} \frac{x}{5}\right]_{0}^{1}\)
\(=\frac{1}{9}\left[\frac{1}{4} \tan ^{-1}\left(\frac{1}{4}\right)-\frac{1}{5} \tan ^{-1}\left(\frac{1}{5}\right)\right]\)
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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.
