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- Improper integral
- Double Integral
- Triple Integral
- Curvilinear integral
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How to use it?
- Integral of d{x}:
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Integral of xcos^2x
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Integral of sin(x/2)*cos(x/2)
- Integral of (4x^2+12x+9)^1/3
- Integral of sin(πt/2)
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Identical expressions
- sqrt((3x- four)(3x+ four))
- square root of ((3x minus 4)(3x plus 4))
- square root of ((3x minus four)(3x plus four))
- √((3x-4)(3x+4))
- sqrt3x-43x+4
- sqrt((3x-4)(3x+4))dx
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Similar expressions
- sqrt((3x-4)(3x-4))
- sqrt((3x+4)(3x+4))
Integral of sqrt((3x-4)(3x+4)) dx
The solution
You have entered
[src]
1 / | | _____________________ | \/ (3*x - 4)*(3*x + 4) dx | / 0
$$\int\limits_{0}^{1} \sqrt{\left(3 x + 4\right) \left(3 x - 4\right)}\, dx$$
Integral(sqrt((3*x - 1*4)*(3*x + 4)), (x, 0, 1))
Detail solution
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Rewrite the integrand:
SqrtQuadraticRule(a=-16, b=0, c=9, context=sqrt(9*x**2 - 16), symbol=x)
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Add the constant of integration:
The answer is:
The answer (Indefinite)
[src]
/ / ____________ \ ____________ | | / 2 | / 2 | _____________________ 8*log\6*\/ -16 + 9*x + 18*x/ x*\/ -16 + 9*x | \/ (3*x - 4)*(3*x + 4) dx = C - ------------------------------- + ----------------- | 3 2 /
$${{x\,\sqrt{9\,x^2-16}}\over{2}}-{{8\,\log \left(6\,\sqrt{9\,x^2-16}
+18\,x\right)}\over{3}}$$
The graph
The answer
[src]
/ ____\
|\/ 14 |
___ 16*I*asin|------|
I*\/ 7 4*pi*I \ 4 /
------- - ------ + -----------------
2 3 3
$${{i\,\left(16\,\arcsin \left({{3}\over{4}}\right)+3\,\sqrt{7}
\right)}\over{6}}$$
=
=
/ ____\
|\/ 14 |
___ 16*I*asin|------|
I*\/ 7 4*pi*I \ 4 /
------- - ------ + -----------------
2 3 3
$$- \frac{4 i \pi}{3} + \frac{\sqrt{7} i}{2} + \frac{16 i \operatorname{asin}{\left(\frac{\sqrt{14}}{4} \right)}}{3}$$
The graph
Use the examples entering the upper and lower limits of integration.