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Question

Solve:211x(1+x2)dx

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Solution

Given, 211x(1+x2)dx
We can write 1x(1+x2) as 1xx1+x2
211x(1+x2)dx=21(1xx1+x2)dx
Now,
21(1xx1+x2)dx=211xdx21x1+x2dx
So, 211xdx=[ln|x|]21+C1
for, 21x1+x2dx,
Let, u=1+x2dudx=2xdu=2xdx12du=xdx
Substituting these values we get,
21x1+x2dx=2112udu=12211udu=12[ln|u|]21+C2
211xdx21x1+x2dx=[ln|x|]2112[ln|u|]21+C=[ln|x|]2112[ln1+x2]21+C=[ln2ln1]12[ln(1+22)ln(1+12)]+C=[ln2ln1]12[ln5ln2]+C=ln(21)12ln(52)+C211x(1+x2)dx=ln(21)12ln(52)+C.

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