Question

∫2logx dx=?(a)2logx+1(log x+1)+C (b)x(log2+1)(log 2+1)+C(c)2logxlog2+C (d)2logx2+C

Answer 1

❀ $\large{ \underline\textbf{Answer - }}$


∫log(x2)log(x2)+log(x2−12x+36)dx∫log(x2)log(x2)+log(x2−12x+36)dx

Now, log(ab)=blog(a)log(ab)=blog(a)

And, if we factor x2−12x+36x2−12x+36,

1×1=1(coeffofx2),2(−6)=−12(coeff.ofx),−62=36(coeffofx0)1×1=1(coeffofx2),2(−6)=−12(coeff.ofx),−62=36(coeffofx0)

x2−12x+36x2−12x+36

(x−6)2(x−6)2

And, log((x−6)2)=2log(x−6)log((x−6)2)=2log(x−6)

∫2log(x)2log(x)+2log(x−6)dx∫2log(x)2log(x)+2log(x−6)dx

∫2×log(x)2(log(x)+log(x−6)dx∫2×log(x)2(log(x)+log(x−6)dx

22  and 22  cancel,

∫log(x)log(x)+log(x−6)dx∫log(x)log(x)+log(x−6)dx

Also, log(a)+log(b)=log(ab)log(a)+log(b)=log(ab)

∫log(x)log(x2−6x)dx∫log(x)log(x2−6x)dx

This could be calculated by this rule called Integration by Parts,

∫u(x)v′(x)dx=u(x)v(x)−∫u′(x)v(x)dx∫u(x)v′(x)dx=u(x)v(x)−∫u′(x)v(x)dx

If u(x)=log(x),v′(x)=1log(x2−6x),v(x)=∫1log(x2−6x)u(x)=log(x),v′(x)=1log(x2−6x),v(x)=∫1log(x2−6x)

So,

∫log(x)log(x2−6x)dx∫log(x)log(x2−6x)dx

log(x)×(∫1ln(x2+6x)dx)−∫1x×(∫1ln(x2+6x)dxdxlog(x)×(∫1ln(x2+6x)dx)−∫1x×(∫1ln(x2+6x)dxdx

This is complex, we cant find it

Answer 2

∫log(x2)log(x2)+log(x2−12x+36)dx∫log(x2)log(x2)+log(x2−12x+36)dx

Now, log(ab)=blog(a)log(ab)=blog(a)

And, if we factor x2−12x+36x2−12x+36,

1×1=1(coeffofx2),2(−6)=−12(coeff.ofx),−62=36(coeffofx0)1×1=1(coeffofx2),2(−6)=−12(coeff.ofx),−62=36(coeffofx0)

x2−12x+36x2−12x+36

(x−6)2(x−6)2

And, log((x−6)2)=2log(x−6)log((x−6)2)=2log(x−6)

∫2log(x)2log(x)+2log(x−6)dx∫2log(x)2log(x)+2log(x−6)dx

∫2×log(x)2(log(x)+log(x−6)dx∫2×log(x)2(log(x)+log(x−6)dx

22  and 22  cancel,

∫log(x)log(x)+log(x−6)dx∫log(x)log(x)+log(x−6)dx

Also, log(a)+log(b)=log(ab)log(a)+log(b)=log(ab)

∫log(x)log(x2−6x)dx∫log(x)log(x2−6x)dx

This could be calculated by this rule called Integration by Parts,

∫u(x)v′(x)dx=u(x)v(x)−∫u′(x)v(x)dx∫u(x)v′(x)dx=u(x)v(x)−∫u′(x)v(x)dx

If u(x)=log(x),v′(x)=1log(x2−6x),v(x)=∫1log(x2−6x)u(x)=log(x),v′(x)=1log(x2−6x),v(x)=∫1log(x2−6x)

So,

∫log(x)log(x2−6x)dx∫log(x)log(x2−6x)dx

log(x)×(∫1ln(x2+6x)dx)−∫1x×(∫1ln(x2+6x)dxdxlog(x)×(∫1ln(x2+6x)dx)−∫1x×(∫1ln(x2+6x)dxdx