Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_1^e \frac{e^x}{x}\left( 1 + x \log x \right)\ d\ x\ . Then, \]
\[I = \int_1^e \left( \frac{e^x}{x} + e^x \log x \right) dx\]
\[ \Rightarrow I = \int_1^e \frac{e^x}{x} dx + \int_1^e e^x \log x\ d\ x\]
\[\text{Integrating first term by parts}\]
\[ \Rightarrow I = \left[ \log x e^x \right]_1^e - \int_1^e e^x \log x d x + \int_1^e e^x \log\ x\ d\ x\]
\[ \Rightarrow I = \left( \log e \right) e^e - 0\]
\[ \Rightarrow I = e^e\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
If f(x) is a continuous function defined on [−a, a], then prove that
Evaluate each of the following integral:
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
Choose the correct alternative:
Γ(n) is
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
