Evaluate the following integrals. \int_{0}^{1}2x\tan^{-1}(x^{2})dx

floymdiT

floymdiT

Answered question

2021-10-11

Evaluate the following integrals.
012xtan1(x2)dx

Answer & Explanation

gotovub

gotovub

Skilled2021-10-12Added 98 answers

Here, 012xtan1(x2)dx
Let t=x2. So, dt = 2xdx
Also, when x=0t=0 and x=1t=1
So,
012xtan1(x2)dx
=01tan1tdt
Step 2
By the method of by parts,
=01tan1tdt
=[ttan1t]0101tddt(tan1t)dt
=(π40)01t×11+t2dt
=π401t1+t2dt
=π4[12ln(1+t2)]01
=π4(12ln20)
=π412ln2

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