Integral of sqrt(4x+13) dx
The solution
Detail solution
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Let u=4x+13.
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+13)23
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Now simplify:
6(4x+13)23
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Add the constant of integration:
6(4x+13)23+constant
The answer is:
6(4x+13)23+constant
The answer (Indefinite)
[src]
/
| 3/2
| __________ (4*x + 13)
| \/ 4*x + 13 dx = C + -------------
| 6
/
∫4x+13dx=C+6(4x+13)23
The graph
____ ____
13*\/ 13 17*\/ 17
- --------- + ---------
6 6
−61313+61717
=
____ ____
13*\/ 13 17*\/ 17
- --------- + ---------
6 6
−61313+61717
-13*sqrt(13)/6 + 17*sqrt(17)/6
Use the examples entering the upper and lower limits of integration.