How do you find the antiderivative of #int x(x^2+1)^100 dx#?

Question

Isabella Ackerman · Accepted Answer

#(1/202)(x^2+1)^101+C#

Write down #(x^2+1)^101+C# as a guess and differentiate it (chain rule). You get #(101).(x^2+1)^(101-1).(2x)=202(x^2+1)^100#. This is close, but too big by a factor of #202#. so divide the first guess by #202#.

Alternatively, substitute #u=x^2+1#, giving #(du)/(dx)=2x#, #dx/(du)=1/(2x)#.

Then by the substitution formula the integral becomes #int cancel(x).u^100.(1/(2 cancel(x)))du# #=(1/2)intu^100du# #=(1/2)(1/(100+1))u^(100+1)+C# by the power law #=(1/202)(x^2+1)^101+C#

Whatever you do, don't try to expand the bracket by the binomial expansion!

Elijah Abram · Answer

undefined To find the antiderivative of $ \int x(x^2+1)^{100} \, dx $, you can use the substitution method. Let $ u = x^2 + 1 $. Then, $ du = 2x \, dx $. Rearrange this to express $ dx $ in terms of $ du $, giving $ dx = \frac{1}{2x} \, du $. Substitute these expressions into the integral:

$$ \int x(x^2+1)^{100} \, dx = \int x u^{100} \cdot \frac{1}{2x} \, du $$

Simplify:

$$ \frac{1}{2} \int u^{100} \, du $$

Now integrate $ u^{100} $ with respect to $ u $:

$$ \frac{1}{2} \cdot \frac{u^{101}}{101} + C $$

Substitute back $ u = x^2 + 1 $:

$$ \frac{1}{202} (x^2 + 1)^{101} + C $$

So, the antiderivative of $ \int x(x^2+1)^{100} \, dx $ is $ \frac{1}{202} (x^2 + 1)^{101} + C $, where $ C $ is the constant of integration.