Advertisements
Advertisements
Question
Advertisements
Solution
\[\text{We have}, \]
\[I = \int_0^1 \frac{2x}{1 + x^2} d x\]
\[\text{Putting} 1 + x^2 = t\]
\[ \Rightarrow 2x\ dx = dt\]
\[\text{When } x \to 0; t \to 1\]
\[\text{And } x \to 1; t \to 2\]
\[ \therefore I = \int_1^2 \frac{d t}{t}\]
\[ = \left[ \log_e \left| t \right| \right]_1^2 \]
\[ = \log_e 2 - \log_e 1\]
\[ = \log_e 2 - 0\]
\[ = \log_e 2\]
APPEARS IN
RELATED QUESTIONS
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
Evaluate each of the following integral:
If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.
