Integral of x/(sqrt(2x+1)+1) dx
The solution
Detail solution
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Let u=2x+1.
Then let du=2x+1dx and substitute du:
∫u+1u(2u2−21)du
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There are multiple ways to do this integral.
Method #1
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Rewrite the integrand:
u+1u(2u2−21)=2u2−2u
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Integrate term-by-term:
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The integral of a constant times a function is the constant times the integral of the function:
∫2u2du=2∫u2du
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The integral of un is n+1un+1 when n=−1:
∫u2du=3u3
So, the result is: 6u3
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The integral of a constant times a function is the constant times the integral of the function:
∫(−2u)du=−2∫udu
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The integral of un is n+1un+1 when n=−1:
∫udu=2u2
So, the result is: −4u2
The result is: 6u3−4u2
Method #2
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Rewrite the integrand:
u+1u(2u2−21)=2(u+1)u3−2(u+1)u
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Integrate term-by-term:
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The integral of a constant times a function is the constant times the integral of the function:
∫2(u+1)u3du=2∫u+1u3du
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Rewrite the integrand:
u+1u3=u2−u+1−u+11
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Integrate term-by-term:
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The integral of un is n+1un+1 when n=−1:
∫u2du=3u3
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The integral of a constant times a function is the constant times the integral of the function:
∫(−u)du=−∫udu
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The integral of un is n+1un+1 when n=−1:
∫udu=2u2
So, the result is: −2u2
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The integral of a constant is the constant times the variable of integration:
∫1du=u
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The integral of a constant times a function is the constant times the integral of the function:
∫(−u+11)du=−∫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: −log(u+1)
The result is: 3u3−2u2+u−log(u+1)
So, the result is: 6u3−4u2+2u−2log(u+1)
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The integral of a constant times a function is the constant times the integral of the function:
∫(−2(u+1)u)du=−2∫u+1udu
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Rewrite the integrand:
u+1u=1−u+11
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Integrate term-by-term:
-
The integral of a constant is the constant times the variable of integration:
∫1du=u
-
The integral of a constant times a function is the constant times the integral of the function:
∫(−u+11)du=−∫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: −log(u+1)
The result is: u−log(u+1)
So, the result is: −2u+2log(u+1)
The result is: 6u3−4u2
Now substitute u back in:
−2x+6(2x+1)23−41
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Now simplify:
−2x+6(2x+1)23−41
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Add the constant of integration:
−2x+6(2x+1)23−41+constant
The answer is:
−2x+6(2x+1)23−41+constant
The answer (Indefinite)
[src]
/
| 3/2
| x 1 x (2*x + 1)
| --------------- dx = - - + C - - + ------------
| _________ 4 2 6
| \/ 2*x + 1 + 1
|
/
∫2x+1+1xdx=C−2x+6(2x+1)23−41
The graph
___
2 \/ 3
- - + -----
3 2
−32+23
=
___
2 \/ 3
- - + -----
3 2
−32+23
Use the examples entering the upper and lower limits of integration.