Integral of xln(1+x) dx
The solution
Detail solution
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Use integration by parts:
∫udv=uv−∫vdu
Let u(x)=log(x+1) and let dv(x)=x.
Then du(x)=x+11.
To find v(x):
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The integral of xn is n+1xn+1 when n=−1:
∫xdx=2x2
Now evaluate the sub-integral.
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The integral of a constant times a function is the constant times the integral of the function:
∫2(x+1)x2dx=2∫x+1x2dx
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Rewrite the integrand:
x+1x2=x−1+x+11
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Integrate term-by-term:
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The integral of xn is n+1xn+1 when n=−1:
∫xdx=2x2
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The integral of a constant is the constant times the variable of integration:
∫(−1)dx=−x
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Let u=x+1.
Then let du=dx and substitute du:
∫u1du
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The integral of u1 is log(u).
Now substitute u back in:
log(x+1)
The result is: 2x2−x+log(x+1)
So, the result is: 4x2−2x+2log(x+1)
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Add the constant of integration:
2x2log(x+1)−4x2+2x−2log(x+1)+constant
The answer is:
2x2log(x+1)−4x2+2x−2log(x+1)+constant
The answer (Indefinite)
[src]
/ 2 2
| x log(1 + x) x x *log(1 + x)
| x*log(1 + x) dx = C + - - ---------- - -- + -------------
| 2 2 4 2
/
∫xlog(x+1)dx=C+2x2log(x+1)−4x2+2x−2log(x+1)
The graph
Use the examples entering the upper and lower limits of integration.