\int(x^{2}+1) \sqrt{x+1} d x is equal to

Correct Answer is : 2[(x+1)7/27−2(x+1)5/25+2(x+1)3/23]+c

{(x+1)^{7{/}2}}/{7}-2 {(x+1)^{5 {/}2}}/{5} +2 {(x+1)^{3 {/} 2}}/{3}+c

2[ {(x+1)^{7{/}2}}/{7}-2 {(x+1)^{5 {/} 2}}/{5}+2 {(x+1)^{3{ /} 2}}/{3}]+c

{(x+1)^{7 {/} 2}}/{7}-2 {(x+1)^{5{/}2}}/{5}+c

{(x+1)^{7 {/}2}}/{7}-3 {(x+1)^{5{/} 2}}/{5}+11(x+1)^{1 {/}2}+c

(x+1)^{y{/}2}+(x+1)^{5{/}2}+(x+1)^{3{/}2}+c

KEAM 2010 Math Paper

Question : 44 of 120

Marks: +1, -0

∫(x2+1)√x+1dx is equal to

‌
(x+1)7/2
7
−2‌
(x+1)5/2
5
+2‌
(x+1)3/2
3
+c
2[‌
(x+1)7/2
7
−2‌
(x+1)5/2
5
+2‌
(x+1)3/2
3
]+c
‌
(x+1)7/2
7
−2‌
(x+1)5/2
5
+c
‌
(x+1)7/2
7
−3‌
(x+1)5/2
5
+11(x+1)1/2+c
(x+1)y/2+(x+1)5/2+(x+1)3/2+c

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