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- Improper integral
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How to use it?
- Integral of d{x}:
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Integral of 2
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Integral of dx
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Integral of e^(2*x)
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Integral of e^(3x)
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Identical expressions
- (4x- three)/(two x^2-3x- seven)
- (4x minus 3) divide by (2x squared minus 3x minus 7)
- (4x minus three) divide by (two x squared minus 3x minus seven)
- (4x-3)/(2x2-3x-7)
- 4x-3/2x2-3x-7
- (4x-3)/(2x²-3x-7)
- (4x-3)/(2x to the power of 2-3x-7)
- 4x-3/2x^2-3x-7
- (4x-3) divide by (2x^2-3x-7)
- (4x-3)/(2x^2-3x-7)dx
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Similar expressions
- (4x-3)/(2x^2+3x-7)
- (4x+3)/(2x^2-3x-7)
- (4x-3)/(2x^2-3x+7)
Integral of (4x-3)/(2x^2-3x-7) dx
The solution
You have entered
[src]
1 / | | 4*x - 3 | -------------- dx | 2 | 2*x - 3*x - 7 | / 0
$$\int\limits_{0}^{1} \frac{4 x - 3}{2 x^{2} - 3 x - 7}\, dx$$
Detail solution
We have the integral:
/ | | 4*x - 3 | 1*-------------- dx | 2 | 2*x - 3*x - 7 | /
Rewrite the integrand
4*x - 3 2*2*x - 3 -------------- = 1*-------------- 2 2 2*x - 3*x - 7 2*x - 3*x - 7
or
/ | | 4*x - 3 | 1*-------------- dx | 2 = | 2*x - 3*x - 7 | /
/ | | 2*2*x - 3 | -------------- dx | 2 | 2*x - 3*x - 7 | /
In the integral
/ | | 2*2*x - 3 | -------------- dx | 2 | 2*x - 3*x - 7 | /
do replacement
2 u = -3*x + 2*x
then
the integral =
/ | | 1 | ------ du = log(-7 + u) | -7 + u | /
do backward replacement
/ | | 2*2*x - 3 / 2\ | -------------- dx = log\-7 - 3*x + 2*x / | 2 | 2*x - 3*x - 7 | /
Solution is:
/ 2\ C + log\-7 - 3*x + 2*x /
The answer (Indefinite)
[src]
/ | | 4*x - 3 / 2 \ | -------------- dx = C + log\2*x - 3*x - 7/ | 2 | 2*x - 3*x - 7 | /
$$\log \left(2\,x^2-3\,x-7\right)$$
The graph
The answer
[src]
-log(7) + log(8)
$$\log 8-\log 7$$
=
=
-log(7) + log(8)
$$- \log{\left(7 \right)} + \log{\left(8 \right)}$$
The graph
Use the examples entering the upper and lower limits of integration.