Integral of x+lnx dx
The solution
Detail solution
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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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Use integration by parts:
∫udv=uv−∫vdu
Let u(x)=log(x) and let dv(x)=1.
Then du(x)=x1.
To find v(x):
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The integral of a constant is the constant times the variable of integration:
∫1dx=x
Now evaluate the sub-integral.
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The integral of a constant is the constant times the variable of integration:
∫1dx=x
The result is: 2x2+xlog(x)−x
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Now simplify:
2x(x+2log(x)−2)
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Add the constant of integration:
2x(x+2log(x)−2)+constant
The answer is:
2x(x+2log(x)−2)+constant
The answer (Indefinite)
[src]
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| x
| (x + log(x)) dx = C + -- - x + x*log(x)
| 2
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∫(x+log(x))dx=C+2x2+xlog(x)−x
The graph
Use the examples entering the upper and lower limits of integration.