00:01
In this question, we are asked to calculate some integrals.
00:05
And the first integral equals x to the fourth over four minus the integral of the square root of x, which we can rewrite as x to the one -half minus cos x plus c.
00:22
Now the integral of x to the one -half by the power rule equals x to the three -halves divided by three -halves.
00:34
And finally, after simplifying, we'll get x to the fourth over four minus two -thirds x to the three -halves minus cos x plus c.
00:53
Let's move on to the next integral.
00:56
The next integral is the integral of x divided by x squared plus one squared dx.
01:07
Here we'll make a u substitution, u equals x squared plus one.
01:14
Then du equals 2xdx.
01:20
And after the substitution, we'll get the integral of one over u squared, and xdx becomes du over two.
01:29
This equals one -half times the integral of u to the negative second power.
01:36
And this equals one -half times u to the negative first over negative one plus c.
01:45
This equals negative one -half.
01:47
And after remembering that u equals x squared plus one, we'll get multiplied by one over x squared plus one plus c.
01:57
And this equals negative one divided by two times x squared plus one plus c.
02:05
Now the next integral is the integral of the square root of pi, sorry, square root of pi multiplied by the integral of x cos x squared dx...