Integral of -1/(sqrt(3x+4)) dx
The solution
Detail solution
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The integral of a constant times a function is the constant times the integral of the function:
∫(−3x+41)dx=−∫3x+41dx
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Let u=3x+4.
Then let du=23x+43dx and substitute 32du:
∫32du
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The integral of a constant times a function is the constant times the integral of the function:
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The integral of a constant is the constant times the variable of integration:
∫1du=u
So, the result is: 32u
Now substitute u back in:
323x+4
So, the result is: −323x+4
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Now simplify:
−323x+4
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Add the constant of integration:
−323x+4+constant
The answer is:
−323x+4+constant
The answer (Indefinite)
[src]
/
| _________
| -1 2*\/ 3*x + 4
| ----------- dx = C - -------------
| _________ 3
| \/ 3*x + 4
|
/
∫(−3x+41)dx=C−323x+4
The graph
Use the examples entering the upper and lower limits of integration.