integration

∫ tan-1x1+x[lower limit 0 and upper limit 1]= (pi ln2)/k
find k

3 Answers

1
kartick sharma ·

is k = 32

66
kaymant ·

No kartick, that's not correct. The value of k is 8.

11
Tush Watts ·

Use by part,

0 ∫ 1 tan -1 x1+x = 0 ∫ 1 tan -1 x . 11+x

= tan -1 x log (1+x) - 0 ∫ 1 log (1+x) . 11 + x 2

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