sol181121

18-November-2021 Homework:

Compute the following indefinite integrals (Hint: one can use partial fraction decomposition for rational functions). (You must give some necessary steps, and cannot directly write down the answer.)

1. 
2. (x-1)(x%2B3)%7D%20dx#card=math&code=%5Cint%20%5Cfrac%7B2x%5E2%2B1%7D%7B%28x-4%29%28x-1%29%28x%2B3%29%7D%20dx&id=pLLOX)
3. 
4. 
5. %7D%20dx#card=math&code=%5Cint%20%5Cfrac%7B1%7D%7Bx%281%2Bx%5E2%29%7D%20dx&id=uC8bu)

Solution:

1. . So the anti-derivative is .

2. We assume that (x-1)(x%2B3)%7D#card=math&code=%5Cfrac%7BA%7D%7Bx-4%7D%2B%5Cfrac%7BB%7D%7Bx-1%7D%2B%5Cfrac%7BC%7D%7Bx%2B3%7D%20%3D%20%5Cfrac%7B2x%5E2%2B1%7D%7B%28x-4%29%28x-1%29%28x%2B3%29%7D&id=JPFYh). Clearing the denominators, we get
   (x%2B3)%20%2B%20B(x-4)(x%2B3)%2BC(x-4)(x-1)%3D2x%5E2%2B1#card=math&code=A%28x-1%29%28x%2B3%29%20%2B%20B%28x-4%29%28x%2B3%29%2BC%28x-4%29%28x-1%29%3D2x%5E2%2B1&id=AR5uR).
   To determine , we let %22%20aria-hidden%3D%22true%22%3E%0A%20%3Cuse%20xlink%3Ahref%3D%22%23E1-MJMATHI-78%22%20x%3D%220%22%20y%3D%220%22%3E%3C%2Fuse%3E%0A%20%3Cuse%20xlink%3Ahref%3D%22%23E1-MJMAIN-2192%22%20x%3D%22850%22%20y%3D%220%22%3E%3C%2Fuse%3E%0A%20%3Cuse%20xlink%3Ahref%3D%22%23E1-MJMAIN-34%22%20x%3D%222128%22%20y%3D%220%22%3E%3C%2Fuse%3E%0A%3C%2Fg%3E%0A%3C%2Fsvg%3E#card=math&code=x%5Cto%204&id=OwdfJ), then LHS , RHS . So .
   Similarly, we get that .
   So the anti-derivative is .

3. Observe that (2x%2B1)#card=math&code=2x%5E3%2B3x%5E2%20%2B%20x%20%3D%20x%28x%2B1%29%282x%2B1%29&id=K7p8H).
   The partial fraction decomposition is .
   The anti-derivative is .

4. We have as the partial fraction decomposition.
   It is straightforward to see that .
   To proceed, we rewrite the second summand as %7D#card=math&code=-%5Cfrac%7B1%7D%7B6%7D%5Cfrac%7B2x-1%7D%7Bx%5E2-x%2B1%7D%2B%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B1%7D%7B%28x%5E2-x%2B1%29%7D&id=EfKPo).
   Since the discriminant of is . So we calculate it as follows:

%5E2%2B%5Cfrac%7B3%7D%7B4%7D%7D%20%3D%20%5Cfrac%7B2%7D%7B3%7D%5Cfrac%7B1%7D%7B1%2B%5Cbig(%5Cfrac%7B2x-1%7D%7B%5Csqrt%7B3%7D%7D%5Cbig)%5E2%7D%20%5CRightarrow%20%5Cint%20%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%20x%7D%7B(x-%5Cfrac%7B1%7D%7B2%7D)%5E2%2B%5Cfrac%7B3%7D%7B4%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D%5Carctan%5Cbig(%5Cfrac%7B2x-1%7D%7B%5Csqrt%7B3%7D%7D%5Cbig)%2BC#card=math&code=%5Cdisplaystyle%20%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B1%7D%7B%28x-%5Cfrac%7B1%7D%7B2%7D%29%5E2%2B%5Cfrac%7B3%7D%7B4%7D%7D%20%3D%20%5Cfrac%7B2%7D%7B3%7D%5Cfrac%7B1%7D%7B1%2B%5Cbig%28%5Cfrac%7B2x-1%7D%7B%5Csqrt%7B3%7D%7D%5Cbig%29%5E2%7D%20%5CRightarrow%20%5Cint%20%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%20x%7D%7B%28x-%5Cfrac%7B1%7D%7B2%7D%29%5E2%2B%5Cfrac%7B3%7D%7B4%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D%5Carctan%5Cbig%28%5Cfrac%7B2x-1%7D%7B%5Csqrt%7B3%7D%7D%5Cbig%29%2BC&id=H6V7m).
So, the original indefinite-integral is equal to
%2B%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D%5Carctan%5Cbig(%5Cfrac%7B2x-1%7D%7B%5Csqrt%7B3%7D%7D%5Cbig)%2BC#card=math&code=%5Cfrac%7B1%7D%7B3%7D%5Cln%7Cx%2B1%7C-%5Cfrac%7B1%7D%7B6%7D%5Cln%28x%5E2-x%2B1%29%2B%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D%5Carctan%5Cbig%28%5Cfrac%7B2x-1%7D%7B%5Csqrt%7B3%7D%7D%5Cbig%29%2BC&id=MKEVP).

1. It is easy to see that %7D%3D%5Cfrac%7B1%7D%7Bx%7D-%5Cfrac%7Bx%7D%7Bx%5E2%2B1%7D#card=math&code=%5Cfrac%7B1%7D%7Bx%28x%5E2%2B1%29%7D%3D%5Cfrac%7B1%7D%7Bx%7D-%5Cfrac%7Bx%7D%7Bx%5E2%2B1%7D&id=N63zI). So the anti-derivative is %2BC#card=math&code=%5Cln%7Cx%7C-%5Cfrac%7B1%7D%7B2%7D%5Cln%28x%5E2%2B1%29%2BC&id=Gp7aM).