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- Improper integral
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How to use it?
- Integral of d{x}:
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Integral of cosx
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Integral of (1+4x^2)^(1/2)
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Integral of (x^2)sinx
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Integral of (x^3-17)/(x^2-4x+3)
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Identical expressions
- one /(two x^2+3x- one)
- 1 divide by (2x squared plus 3x minus 1)
- one divide by (two x squared plus 3x minus one)
- 1/(2x2+3x-1)
- 1/2x2+3x-1
- 1/(2x²+3x-1)
- 1/(2x to the power of 2+3x-1)
- 1/2x^2+3x-1
- 1 divide by (2x^2+3x-1)
- 1/(2x^2+3x-1)dx
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Similar expressions
- 1/(2x^2+3x+1)
- 1/(2x^2-3x-1)
Integral of 1/(2x^2+3x-1) dx
The solution
You have entered
[src]
1 / | | 1 | -------------- dx | 2 | 2*x + 3*x - 1 | / 0
$$\int\limits_{0}^{1} \frac{1}{\left(2 x^{2} + 3 x\right) - 1}\, dx$$
Integral(1/(2*x^2 + 3*x - 1), (x, 0, 1))
The answer (Indefinite)
[src]
// / ____ \ \
|| ____ |4*\/ 17 *(3/4 + x)| |
||-\/ 17 *acoth|------------------| |
/ || \ 17 / 2 17|
| ||---------------------------------- for (3/4 + x) > --|
| 1 || 68 16|
| -------------- dx = C + 8*|< |
| 2 || / ____ \ |
| 2*x + 3*x - 1 || ____ |4*\/ 17 *(3/4 + x)| |
| ||-\/ 17 *atanh|------------------| |
/ || \ 17 / 2 17|
||---------------------------------- for (3/4 + x) < --|
\\ 68 16/
$$\int \frac{1}{\left(2 x^{2} + 3 x\right) - 1}\, dx = C + 8 \left(\begin{cases} - \frac{\sqrt{17} \operatorname{acoth}{\left(\frac{4 \sqrt{17} \left(x + \frac{3}{4}\right)}{17} \right)}}{68} & \text{for}\: \left(x + \frac{3}{4}\right)^{2} > \frac{17}{16} \\- \frac{\sqrt{17} \operatorname{atanh}{\left(\frac{4 \sqrt{17} \left(x + \frac{3}{4}\right)}{17} \right)}}{68} & \text{for}\: \left(x + \frac{3}{4}\right)^{2} < \frac{17}{16} \end{cases}\right)$$
The graph
Use the examples entering the upper and lower limits of integration.