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int(x^(2)+1)sqrt(x+1)dx is equal to...

`int(x^(2)+1)sqrt(x+1)dx` is equal to

A

`((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`

Text Solution

Verified by Experts

The correct Answer is:
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`
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