If ∫(x)5(x)7+x6dx=alnxkxk+1+c, the value of a and k, respectively, are
52 and 25
25 and 52
52 and 2
None of these
∫dx(x)2+(x)7=∫dx(x)71+1(x)5
Put 1(x)5=y,dydx=−52(x)7
∴I=∫−2dy5(1+y)=−25ln|1+y|+c=25ln11+1(x)5∴a=25,k=52