If f(x)=∫5x8+7x6x2+1+2x72dx, and f(0)=0, then the value  of f(1) is

f(x)=∫5x8+7x6x142+1x7+1x52dx=∫5x6+7x82+1x7+1x52dx Put, 2+1x7+1x5=t⇒ f(x)=−∫dtt2=1t+c=x72x7+x2+1+cf(0)=0⇒ c=0⇒ f(x)=x72x7+x2+1⇒ f(1)=14