`int(x^(2)+1)sqrt(x+1)dx` is equal toA. `((x+1)^(7//2))/(7)-2((x+1)^(5//2))/(5)+2((x+1)^(3//2))/(3)+C`B. `2[((x+1)^(7//2))/(7)-2((x+1)^(5//2))/(5)+2((x+1)^(3//2))/(3)]+C`C. `((x+1)^(7//2))/(5)-2((x+1)^(5//2))/(5)+5`D. `((x+1)^(7//2))/(7)-3((x+1)^(5//2))/(5)+11(x+1)^(1//2)+C`

`int(x^(2)+1)sqrt(x+1)dx` is equal toA. `((x+1)^(7//2))/(7)-2((x+1)^(5

1.	`int(x^(2)+1)sqrt(x+1)dx` is equal toA. `((x+1)^(7//2))/(7)-2((x+1)^(5//2))/(5)+2((x+1)^(3//2))/(3)+C`B. `2[((x+1)^(7//2))/(7)-2((x+1)^(5//2))/(5)+2((x+1)^(3//2))/(3)]+C`C. `((x+1)^(7//2))/(5)-2((x+1)^(5//2))/(5)+5`D. `((x+1)^(7//2))/(7)-3((x+1)^(5//2))/(5)+11(x+1)^(1//2)+C`
Answer» Correct Answer - B Put `x+1=t^(2)` `rArr" "dx=2t dt` `rArr" "x^(2)+1=(t^(2)-1)^(2)+1=t^(4)-2t^(2)+2` `therefore" Given integral "=int(t^(4)-2t^(2)+2)t.2t dt` `=2int(t^(6)-2t^(4)+2t^(2))dt` `=2[(t^(7))/(7)-2(t^(5))/(5)+2(t^(3))/(3)]+C` `=2[(x+1)^((7)/(2))/(7)-2(x+1)^((5)/(2))/(5)+2(x+1)^((3)/(2))/(3)]+C`