Question

EVALUATE............

Answer 1

Differentiating both sides,*

−2x3dx=dt or dxx3=−12dt *

∴I=−12∫t√logtdt=−12∫(logt)⋅t1/2dt*

*

Integrating by Product Rule, =*

−12[logt)⋅t3/23/2−∫1t⋅t3/23/2dt]*

=−13(nlogt+13∫t1/2dt *

=−13t3nlogt+13⋅t3/23/2+c*

=29t3n−13t2nlogt+c=13tn[23−logt]+c *

Putting t=1+1x2, *

we have =13(1+1x2)3/2[23−log(1+1x2)]+c