Question:
$ \int_{-1}^{1}{\frac{17{{x}^{5}}-{{x}^{4}}+29{{x}^{3}}-31x+1}{{{x}^{2}}+1}}dx $ is
$ \int_{-1}^{1}{\frac{17{{x}^{5}}-{{x}^{4}}+29{{x}^{3}}-31x+1}{{{x}^{2}}+1}}dx $ is
Updated On: May 11, 2024
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The Correct Option is C
Solution and Explanation
Let $ I=\int_{-1}^{1}{\frac{17{{x}^{5}}-{{x}^{4}}+28{{x}^{3}}-31x+1}{{{x}^{2}}+1}}dx $ $ I=\int_{-1}^{1}{\frac{17{{x}^{5}}+29{{x}^{3}}-31x}{{{x}^{2}}+1}}dx-\int_{-1}^{1}{\frac{{{x}^{4}}-1}{{{x}^{2}}+1}}dx $
$=-2\int_{0}^{1}{\frac{({{x}^{2}}-1)({{x}^{2}}+1)}{({{x}^{2}}+1)}}dx $
$=2\left[ \left( \frac{{{x}^{3}}}{3}-x \right) \right]_{0}^{1}=-2\left[ \frac{1}{3}-1 \right] $
$=\frac{4}{3} $
$=-2\int_{0}^{1}{\frac{({{x}^{2}}-1)({{x}^{2}}+1)}{({{x}^{2}}+1)}}dx $
$=2\left[ \left( \frac{{{x}^{3}}}{3}-x \right) \right]_{0}^{1}=-2\left[ \frac{1}{3}-1 \right] $
$=\frac{4}{3} $
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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.
