Integral of 1/(sqrt(3*x+1)-1) dx
The solution
Detail solution
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Let u=3x+1.
Then let du=23x+13dx and substitute 2du:
∫3u−32udu
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The integral of a constant times a function is the constant times the integral of the function:
∫3u−3udu=2∫3u−3udu
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Rewrite the integrand:
3u−3u=31+3(u−1)1
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Integrate term-by-term:
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The integral of a constant is the constant times the variable of integration:
∫31du=3u
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The integral of a constant times a function is the constant times the integral of the function:
∫3(u−1)1du=3∫u−11du
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Let u=u−1.
Then let du=du and substitute du:
∫u1du
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The integral of u1 is log(u).
Now substitute u back in:
log(u−1)
So, the result is: 3log(u−1)
The result is: 3u+3log(u−1)
So, the result is: 32u+32log(u−1)
Now substitute u back in:
323x+1+32log(3x+1−1)
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Now simplify:
323x+1+32log(3x+1−1)
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Add the constant of integration:
323x+1+32log(3x+1−1)+constant
The answer is:
323x+1+32log(3x+1−1)+constant
The answer (Indefinite)
[src]
/
| _________ / _________\
| 1 2*\/ 3*x + 1 2*log\-1 + \/ 3*x + 1 /
| --------------- dx = C + ------------- + -----------------------
| _________ 3 3
| \/ 3*x + 1 - 1
|
/
∫3x+1−11dx=C+323x+1+32log(3x+1−1)
The graph
Use the examples entering the upper and lower limits of integration.