(i) Evaluate ∫ x/x^2+1 dx ,x∈[2,3] (ii) Evaluate ∫ x/1+sinx dx x∈[0,π]

Question

(i) Evaluate:  \(\int\limits_2^3\) \(\frac{x}{x^2+1}\)dx

(ii) Evaluate :\(\int\limits_0^π\)\(\frac{x}{1+sinx}dx\)

Answer 1

= π [tanπ - secπ - tan0 + sec0]

= π [0-(-1) -0 +1] = 2π

⇒ I = π