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