Integral of dx/(sqrt(3)*sqrt(x)+4) dx
The solution
Detail solution
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Let u=x.
Then let du=2xdx and substitute 2du:
∫3u+42udu
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The integral of a constant times a function is the constant times the integral of the function:
∫3u+4udu=2∫3u+4udu
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Rewrite the integrand:
3u+4u=33−3(3u+4)43
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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:
∫33du=33u
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The integral of a constant times a function is the constant times the integral of the function:
∫(−3(3u+4)43)du=−343∫3u+41du
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Let u=3u+4.
Then let du=3du and substitute 33du:
∫3u3du
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The integral of a constant times a function is the constant times the integral of the function:
∫u1du=33∫u1du
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The integral of u1 is log(u).
So, the result is: 33log(u)
Now substitute u back in:
33log(3u+4)
So, the result is: −34log(3u+4)
The result is: 33u−34log(3u+4)
So, the result is: 323u−38log(3u+4)
Now substitute u back in:
323x−38log(3x+4)
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Add the constant of integration:
323x−38log(3x+4)+constant
The answer is:
323x−38log(3x+4)+constant
The answer (Indefinite)
[src]
/
| / ___ ___\ ___ ___
| 1 8*log\4 + \/ 3 *\/ x / 2*\/ 3 *\/ x
| --------------- dx = C - ---------------------- + -------------
| ___ ___ 3 3
| \/ 3 *\/ x + 4
|
/
∫3x+41dx=C+323x−38log(3x+4)
The graph
/ ___\ ___ / ___\ ___
8*log\4 + 2*\/ 3 / 4*\/ 3 8*log\4 + I*\/ 3 / 2*I*\/ 3
- ------------------ + ------- + ------------------ - ---------
3 3 3 3
−38log(23+4)+343−323i+38log(4+3i)
=
/ ___\ ___ / ___\ ___
8*log\4 + 2*\/ 3 / 4*\/ 3 8*log\4 + I*\/ 3 / 2*I*\/ 3
- ------------------ + ------- + ------------------ - ---------
3 3 3 3
−38log(23+4)+343−323i+38log(4+3i)
-8*log(4 + 2*sqrt(3))/3 + 4*sqrt(3)/3 + 8*log(4 + i*sqrt(3))/3 - 2*i*sqrt(3)/3
(0.87495726530422 - 0.0650215499529942j)
(0.87495726530422 - 0.0650215499529942j)
Use the examples entering the upper and lower limits of integration.