Advertisements
Advertisements
Question
Advertisements
Solution
\[Let\ I = \int_1^e \frac{e^x}{x}\left( 1 + x \log x \right)\ d\ x\ . Then, \]
\[I = \int_1^e \left( \frac{e^x}{x} + e^x \log x \right) dx\]
\[ \Rightarrow I = \int_1^e \frac{e^x}{x} dx + \int_1^e e^x \log x\ d\ x\]
\[\text{Integrating first term by parts}\]
\[ \Rightarrow I = \left[ \log x e^x \right]_1^e - \int_1^e e^x \log x d x + \int_1^e e^x \log\ x\ d\ x\]
\[ \Rightarrow I = \left( \log e \right) e^e - 0\]
\[ \Rightarrow I = e^e\]
APPEARS IN
RELATED QUESTIONS
Write the coefficient a, b, c of which the value of the integral
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
Find : `∫_a^b logx/x` dx
Evaluate the following:
`int_0^2 "f"(x) "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`
Evaluate the following using properties of definite integral:
`int_(- pi/4)^(pi/4) x^3 cos^3 x "d"x`
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
Find: `int logx/(1 + log x)^2 dx`
