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- Improper integral
- Double Integral
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How to use it?
- Integral of d{x}:
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Integral of e^3x
- Integral of sinx/cos2x
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Integral of -e^x
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Integral of sqrt(x^2+1)/x^3
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Identical expressions
- (x- one)/((3x- four)^ two)+ one
- (x minus 1) divide by ((3x minus 4) squared ) plus 1
- (x minus one) divide by ((3x minus four) to the power of two) plus one
- (x-1)/((3x-4)2)+1
- x-1/3x-42+1
- (x-1)/((3x-4)²)+1
- (x-1)/((3x-4) to the power of 2)+1
- x-1/3x-4^2+1
- (x-1) divide by ((3x-4)^2)+1
- (x-1)/((3x-4)^2)+1dx
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Similar expressions
- (x-1)/((3x-4)^2)-1
- (x-1)/((3x+4)^2)+1
- (x+1)/((3x-4)^2)+1
Integral of (x-1)/((3x-4)^2)+1 dx
The solution
You have entered
[src]
4 / | | / x - 1 \ | |---------- + 1| dx | | 2 | | \(3*x - 4) / | / 1
$$\int\limits_{1}^{4} \left(\frac{x - 1}{\left(3 x - 4\right)^{2}} + 1\right)\, dx$$
Integral((x - 1)/(3*x - 4)^2 + 1, (x, 1, 4))
The answer (Indefinite)
[src]
/ | | / x - 1 \ 1 log(-4 + 3*x) | |---------- + 1| dx = C + x - ------------ + ------------- | | 2 | 9*(-4 + 3*x) 9 | \(3*x - 4) / | /
$$\int \left(\frac{x - 1}{\left(3 x - 4\right)^{2}} + 1\right)\, dx = C + x + \frac{\log{\left(3 x - 4 \right)}}{9} - \frac{1}{9 \left(3 x - 4\right)}$$
The graph
The answer
[src]
pi*I
oo - ----
9
$$\infty - \frac{i \pi}{9}$$
=
=
pi*I
oo - ----
9
$$\infty - \frac{i \pi}{9}$$
oo - pi*i/9
Use the examples entering the upper and lower limits of integration.