Advertisements
Advertisements
Question
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
Advertisements
Solution
Let v = b2 + c2x2, then dv = 2c2 xdx
Therefore, `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
= `(3"a")/(2"c"^2) int "dv"/"v"`
= `(3"a")/("c"^2) log |"b"^2 + "c"^2x^2| + "C"`
APPEARS IN
RELATED QUESTIONS
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
Evaluate the following definite integrals:
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
Solve each of the following integral:
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`
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_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
The value of `int_2^3 x/(x^2 + 1)`dx is ______.
