Advertisements
Advertisements
Question
Advertisements
Solution
\[I = \int_0^1 \frac{2x}{5 x^2 + 1} d x + \int_0^1 \frac{3}{5 x^2 + 1} d x\]
\[ \Rightarrow I = \frac{1}{5} \int_0^1 \frac{10x}{5 x^2 + 1} d x + 3 \int_0^1 \frac{1}{\left( \sqrt{5}x \right)^2 + 1^2} d x\]
\[ \Rightarrow I = \frac{1}{5} \left[ \log \left( 5 x^2 + 1 \right) \right]_0^1 + \frac{3}{\sqrt{5}} \left[ \tan^{- 1} \left( \sqrt{5}x \right) \right]_0^1 \]
\[ \Rightarrow I = \frac{1}{5} \log 6 + \frac{3}{\sqrt{5}} \tan^{- 1} \sqrt{5}\]
APPEARS IN
RELATED QUESTIONS
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
Evaluate each of the following integral:
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
\[\int\limits_0^4 x dx\]
Evaluate the following using properties of definite integral:
`int_(- pi/4)^(pi/4) x^3 cos^3 x "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
If velocity changes with time, what does a definite integral give over a time interval?
