∫x(1+x)32dx=?
2(2+x)(1+x)+c
(2+x)(1+x)+c
32x(1-x)+c
32(2+x)(1+x)+c
Explanation For The Correct Option:
Evaluating the integral
∫x(1+x)32dx
Substituting 1+x=t2
Differentiating it w.r.t x then dx=2tdt
∫x1+x32=∫t2-1(2tdt)t232=2∫tt2-1t3dt=2∫1-1t2dt=2∫1-t-2dt=2t+t-1∵∫xndx=xn+1n+1+c=2t+1t=2t2+1t+c[c=constant]=2x+21+x+c[substitutingback1+x=t2]
Hence, the correct answer is option (A).