Question:
$\int \frac{e^{x}}{x}\left(x\,log\,x+1\right)dx$ is equal to
$\int \frac{e^{x}}{x}\left(x\,log\,x+1\right)dx$ is equal to
Updated On: Apr 8, 2024
- $\frac{e^{x}}{x}+C$
- $xe^x\, log \,| x | +C$
- $e^x\, log | x | +C$
- $x(e^x +log \,|x|) + C$
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The Correct Option is C
Solution and Explanation
$\int \frac{e^{x}}{x}(x \log x+1) d x$
$=\int \underset{II}{e^{x}} \underset{I}{\log} x d x+\int \frac{e^{x}}{x} d x$
$=e^{x} \log x-\int \frac{e^{x}}{x} d x+\int \frac{e^{x}}{x} d x+C$
$=e^{x} \log x+C$
$=\int \underset{II}{e^{x}} \underset{I}{\log} x d x+\int \frac{e^{x}}{x} d x$
$=e^{x} \log x-\int \frac{e^{x}}{x} d x+\int \frac{e^{x}}{x} d x+C$
$=e^{x} \log x+C$
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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.
