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How to use it?
- Integral of d{x}:
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Integral of (4-x^2)^(1/2)
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Integral of (x+2)e^x
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Integral of 1/cos^2(x)
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Integral of x^3/(x^2+1)
- How do you in partial fractions?:
- (3-4x)/(2x^2-3x+1)
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Identical expressions
- (three -4x)/(two x^2-3x+ one)
- (3 minus 4x) divide by (2x squared minus 3x plus 1)
- (three minus 4x) divide by (two x squared minus 3x plus one)
- (3-4x)/(2x2-3x+1)
- 3-4x/2x2-3x+1
- (3-4x)/(2x²-3x+1)
- (3-4x)/(2x to the power of 2-3x+1)
- 3-4x/2x^2-3x+1
- (3-4x) divide by (2x^2-3x+1)
- (3-4x)/(2x^2-3x+1)dx
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Similar expressions
- (3+4x)/(2x^2-3x+1)
- (3-4x)/(2x^2+3x+1)
- (3-4x)/(2x^2-3x-1)
Integral of (3-4x)/(2x^2-3x+1) dx
The solution
You have entered
[src]
1 / | | 3 - 4*x | -------------- dx | 2 | 2*x - 3*x + 1 | / 0
$$\int\limits_{0}^{1} \frac{3 - 4 x}{2 x^{2} - 3 x + 1}\, dx$$
Integral((3 - 4*x)/(2*x^2 - 3*x + 1), (x, 0, 1))
Detail solution
We have the integral:
/ | | 3 - 4*x | 1*-------------- dx | 2 | 2*x - 3*x + 1 | /
Rewrite the integrand
3 - 4*x 2*2*x - 3 -------------- = - -------------- 2 2 2*x - 3*x + 1 2*x - 3*x + 1
or
/ | | 3 - 4*x | 1*-------------- dx | 2 = | 2*x - 3*x + 1 | /
/ | | 2*2*x - 3 - | -------------- dx | 2 | 2*x - 3*x + 1 | /
In the integral
/ | | 2*2*x - 3 - | -------------- dx | 2 | 2*x - 3*x + 1 | /
do replacement
2 u = -3*x + 2*x
then
the integral =
/ | | 1 - | ----- du = -log(1 + u) | 1 + u | /
do backward replacement
/ | | 2*2*x - 3 / 2\ - | -------------- dx = -log\1 - 3*x + 2*x / | 2 | 2*x - 3*x + 1 | /
Solution is:
/ 2\ C - log\1 - 3*x + 2*x /
The answer (Indefinite)
[src]
/ | | 3 - 4*x / 2 \ | -------------- dx = C - log\2*x - 3*x + 1/ | 2 | 2*x - 3*x + 1 | /
$$-\log \left(2\,x^2-3\,x+1\right)$$
Use the examples entering the upper and lower limits of integration.