Question
Evaluate the integral
21x2ln(x)−41x2+C,C∈R
Evaluate
∫xln(x)dx
Prepare for integration by parts
u=ln(x)dv=xdx
Calculate the derivative
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Calculate the derivative
u=ln(x)
Evaluate the derivative
du=(ln(x))′dx
Evaluate the derivative
du=x1dx
du=x1dxdv=xdx
Evaluate the integral
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Evaluate the integral
dv=xdx
Evaluate the integral
∫1dv=∫xdx
Evaluate the integral
v=21x2
du=x1dxv=21x2
Substitute u=ln(x)、v=21x2、du=x1dx、dv=xdx for ∫udv=uv−∫vdu
ln(x)×21x2−∫x1×21x2dx
Calculate
21x2ln(x)−∫2xdx
Use the property of integral ∫kf(x)dx=k∫f(x)dx
21x2ln(x)−21×∫xdx
Evaluate the integral
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Evaluate the integral
21×∫xdx
Evaluate the integral
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Evaluate
∫xdx
Use ∫xndx=n+1xn+1,n=−1 to evaluate the integral
2x2
Simplify
21x2
21×21x2
Calculate
41x2
21x2ln(x)−41x2
Solution
21x2ln(x)−41x2+C,C∈R
Show Solution