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How to use it?
- Integral of d{x}:
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Integral of x^4*e^x
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Integral of 1/(x^2+4x+8)
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Integral of y/sqrt(1+y^2)
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Integral of (1+cos^2x)/(1+cos2x)
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Identical expressions
- ((two x- three)dx)/(x²-3x+2)
- ((2x minus 3)dx) divide by (x² minus 3x plus 2)
- ((two x minus three)dx) divide by (x² minus 3x plus 2)
- 2x-3dx/x²-3x+2
- ((2x-3)dx) divide by (x²-3x+2)
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Similar expressions
- ((2x-3)dx)/(x²+3x+2)
- ((2x-3)dx)/(x²-3x-2)
- ((2x+3)dx)/(x²-3x+2)
Integral of ((2x-3)dx)/(x²-3x+2) dx
The solution
You have entered
[src]
1 / | | 2*x - 3 | ------------ dx | 2 | x - 3*x + 2 | / 0
$$\int\limits_{0}^{1} \frac{2 x - 3}{\left(x^{2} - 3 x\right) + 2}\, dx$$
Integral((2*x - 3)/(x^2 - 3*x + 2), (x, 0, 1))
Detail solution
We have the integral:
/ | | 2*x - 3 | ------------ dx | 2 | x - 3*x + 2 | /
Rewrite the integrand
True
or
/ | | 2*x - 3 | ------------ dx | 2 = | x - 3*x + 2 | /
/ | | 2*x - 3 | ------------ dx | 2 | x - 3*x + 2 | /
In the integral
/ | | 2*x - 3 | ------------ dx | 2 | x - 3*x + 2 | /
do replacement
2 u = x - 3*x
then
the integral =
/ | | 1 | ----- du = log(2 + u) | 2 + u | /
do backward replacement
/ | | 2*x - 3 / 2 \ | ------------ dx = log\2 + x - 3*x/ | 2 | x - 3*x + 2 | /
Solution is:
/ 2 \ C + log\2 + x - 3*x/
The answer (Indefinite)
[src]
/ | | 2*x - 3 / 2 \ | ------------ dx = C + log\x - 3*x + 2/ | 2 | x - 3*x + 2 | /
$$\int \frac{2 x - 3}{\left(x^{2} - 3 x\right) + 2}\, dx = C + \log{\left(\left(x^{2} - 3 x\right) + 2 \right)}$$
The graph
Use the examples entering the upper and lower limits of integration.