Evaluate:∫(x2+1)(x+1)dx
(x+1)7/27−2(x+1)5/25+2(x+1)3/23+C
2(x+1)7/27−2(x+1)5/25+2(x+1)3/23+C
(x+1)7/27−2(x+1)5/25+5
(x+1)7/27−3(x+1)5/25+1(x+1)1/2+C
(x+1)7/2+(x+1)5/2+(x+1)3/2+C
Evaluating the integral∫(x2+1)(x+1)dx
Substituting t2=(x+1) and dx=2dtwe get
⇒∫(t4−2t2+2)t2tdt∵x2+1=t4−2t2+2⇒2∫(t6−2t4+2t2)dt⇒2∫t6dt-2∫t4+2∫t2dt⇒2t77−2t55+2t33+C⇒2(x+1)727−2(x+1)525+2(x+1)323+C[substitutingbackt2=(x+1)]
Hence, option B is the correct answer.