MCQ
$\int_{}^{} {{e^{2x + \log x}}} dx = $
- ✓$\frac{1}{4}(2x - 1){e^{2x}} + c$
- B$\frac{1}{4}(2x+ 1){e^{2x}} + c$
- C$\frac{1}{2}(2x - 1){e^{2x}} + c$
- D$\frac{1}{2}(2x + 1){e^{2x}} + c$
✓
Answer
Correct option: A.
$\frac{1}{4}(2x - 1){e^{2x}} + c$
a
(a)$\int_{}^{} {{e^{2x + \log x}}dx} = \int_{}^{} {x{e^{2x}}dx} $
$ = \frac{{x{e^{2x}}}}{2} - \int_{}^{} {\frac{1}{2}{e^{2x}}dx + c} = \frac{{{e^{2x}}}}{4}(2x - 1) + c.$
(a)$\int_{}^{} {{e^{2x + \log x}}dx} = \int_{}^{} {x{e^{2x}}dx} $
$ = \frac{{x{e^{2x}}}}{2} - \int_{}^{} {\frac{1}{2}{e^{2x}}dx + c} = \frac{{{e^{2x}}}}{4}(2x - 1) + c.$
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