1. Evaluate the Integral integral of 1/(2x-1) with respect to x - Mathway
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2. How do you integrate 1/(2x+1)? - Socratic
8 jan 2017 · int (dx)/(2x+1) = 1/2 ln abs (2x+1)+C int (dx)/(2x+1) = 1/2 int (d(2x+1))/(2x+1) = 1/2 ln abs (2x+1)+C.
int (dx)/(2x+1) = 1/2 ln abs (2x+1)+C int (dx)/(2x+1) = 1/2 int (d(2x+1))/(2x+1) = 1/2 ln abs (2x+1)+C
3. Find the Antiderivative 1/2x-1 - Mathway
The function F(x) F ( x ) can be found by finding the indefinite integral of the derivative f(x) f ( x ) . F(x)=∫f(x)dx F ( x ) = ∫ f ( x ) d x.
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4. How do you evaluate the integral int 1/(2x-1) from 1 to 2? | Socratic
4 nov 2016 · I found: ln(3)2. Explanation: Try this: enter image source here.
I found: ln(3)/2 Try this:
5. integrate 1/(2x) dx - Wolfram|Alpha
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6. Evaluate displaystyle int { frac { 1 }{(1-2x) } }dx
∫11−2xdxt=1−2x⟹dt=−2dx∫1−21tdt1−2logt+c. Was this answer helpful? upvote 1. Similar Questions. star-struck. Q1. Evaluate ∫1(1−2x)dx. View Solution.
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7. Ex 7.5, 6 - Integrate 1 - x2 / x (1 - 2x) - Class 12 - Teachoo
16 apr 2024 · Ex 7.5, 6 Integrate the function (1 − 𝑥2)/(𝑥(1 − 2𝑥)) ∫1·(1 − 𝑥2)/(𝑥(1 − 2𝑥)) 𝑑𝑥= ∫1·(1 − 𝑥^2)/(𝑥 − 2𝑥^2 ) 𝑑𝑥 =∫1·(1/2 + (− 𝑥/2 + 1)/(−2𝑥^2+ ...
Ex 7.5, 6 Integrate the function (1 − 𝑥2)/(𝑥(1 − 2𝑥)) ∫1▒(1 − 𝑥2)/(𝑥(1 − 2𝑥)) 𝑑𝑥= ∫1▒(1 − 𝑥^2)/(𝑥 − 2𝑥^2 ) 𝑑𝑥 =∫1▒(1/2 + (− 𝑥/2 + 1)/(−2𝑥^2+ 𝑥)) 𝑑𝑥 =1/2 ∫1▒𝑑𝑥+1/2 ∫1▒(−𝑥 + 2)/(−2𝑥^2+ 𝑥) 𝑑𝑥 =1/2 ∫1▒𝑑𝑥−1/2 ∫1▒(𝑥 − 2)/𝑥(1 − 2𝑥) 𝑑𝑥 =1/2 ∫1▒𝑑𝑥−1/2 ∫1▒(𝑥 − 2)/𝑥(
8. Integrate:intfrac{1-x^{2}}{x(1-2x)}
Click here:point_up_2:to get an answer to your question :writing_hand:integrateintfrac1x2x12x.
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9. (i) int(dx)/(2x+1) - Doubtnut
4 mrt 2021 · (i) ∫dx2x+1 ... Step by step video & image solution for (i) int(dx)/(2x+1) by Maths experts to help you in doubts & scoring excellent marks in ...
(i) int(dx)/(2x+1)
10. Integrating 1/2x: Comparing Methods and Solutions - Physics Forums
12 okt 2013 · The process for solving ∫(1/2x)dx is to first write the integral in its proper form and then use the power rule for integration or substitution.
Homework Statement ∫(1/2x)dx The Attempt at a Solution I factor out (1/2) and i get ∫(1/2x)dx=(1/2)ln|x|+C ∫(1/2x)dx (1/2)∫(1/x)dx (1/2)ln|x|+C But can't i say that u=2x and dx=du/2 and get ∫(1/2x)dx= (1/2)ln|2x| + C ∫(1/2x)dx u=2x ∫(1/u)(du/2) ∫(1/2u)du (1/2)∫(1/u)dx...
