Integral of sqrt(4x-5) dx
The solution
Detail solution
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Let u=4x−5.
Then let du=4dx and substitute 4du:
∫4udu
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The integral of a constant times a function is the constant times the integral of the function:
∫udu=4∫udu
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The integral of un is n+1un+1 when n=−1:
∫udu=32u23
So, the result is: 6u23
Now substitute u back in:
6(4x−5)23
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Now simplify:
6(4x−5)23
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Add the constant of integration:
6(4x−5)23+constant
The answer is:
6(4x−5)23+constant
The answer (Indefinite)
[src]
/
| 3/2
| _________ (4*x - 5)
| \/ 4*x - 5 dx = C + ------------
| 6
/
∫4x−5dx=C+6(4x−5)23
The graph
___
I 5*I*\/ 5
- - + ---------
6 6
−6i+655i
=
___
I 5*I*\/ 5
- - + ---------
6 6
−6i+655i
(0.0 + 1.69672331458316j)
(0.0 + 1.69672331458316j)
Use the examples entering the upper and lower limits of integration.