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
If f is an integrable function, show that
Prove that:
Evaluate each of the following integral:
Solve each of the following integral:
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
Write the coefficient a, b, c of which the value of the integral
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\] then the value of I10 + 90I8 is
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
Evaluate the following:
f(x) = `{{:("c"x",", 0 < x < 1),(0",", "otherwise"):}` Find 'c" if `int_0^1 "f"(x) "d"x` = 2
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
