Integral of -sqrt(4x+1) dx
The solution
Detail solution
-
The integral of a constant times a function is the constant times the integral of the function:
∫(−4x+1)dx=−∫4x+1dx
-
Let u=4x+1.
Then let du=4dx and substitute 4du:
∫4udu
-
The integral of a constant times a function is the constant times the integral of the function:
∫udu=4∫udu
-
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+1)23
So, the result is: −6(4x+1)23
-
Now simplify:
−6(4x+1)23
-
Add the constant of integration:
−6(4x+1)23+constant
The answer is:
−6(4x+1)23+constant
The answer (Indefinite)
[src]
/
| 3/2
| _________ (4*x + 1)
| -\/ 4*x + 1 dx = C - ------------
| 6
/
∫(−4x+1)dx=C−6(4x+1)23
The graph
___
1 5*\/ 5
- - -------
6 6
61−655
=
___
1 5*\/ 5
- - -------
6 6
61−655
Use the examples entering the upper and lower limits of integration.