Evaluate Definite Integral $(x-3)^2$ and $(x+4)^2$

[juantheron]

Evaluation of $\displaystyle \int_{-5}^{-7}\ln \left(x-3\right)^2dx+2\int_{0}^{1}\ln(x+4)^2dx$

My Try:: Let $(x-3) = t$ Then $dx = dt$ and changing Limit, we get

and Again put $(x+4) = u,$ Then $dx = du$ and changing Limit, we get

$\displaystyle \int_{-8}^{-10}\ln(t^2)dt+2\int_{4}^{5}\ln(u)^2du$

Now How can I solve after That

Help me

Thanks

 

[kaliprasad]

Evaluation of $\displaystyle \int_{-5}^{-7}\ln \left(x-3\right)^2dx+2\int_{0}^{1}\ln(x+4)^2dx$

My Try:: Let $(x-3) = t$ Then $dx = dt$ and changing Limit, we get

and Again put $(x+4) = u,$ Then $dx = du$ and changing Limit, we get

$\displaystyle \int_{-8}^{-10}\ln(t^2)dt+2\int_{4}^{5}\ln(u)^2du$

Now How can I solve after That

Help me

Thanks

$ln\, t^2 =2 \,ln\, t$ and then I think you know how to integrate ln t dt and as 1st limit is -ve use $(-t)^2$

 

[zzephod]

Evaluation of $\displaystyle \int_{-5}^{-7}\ln \left(x-3\right)^2dx+2\int_{0}^{1}\ln(x+4)^2dx$

My Try:: Let $(x-3) = t$ Then $dx = dt$ and changing Limit, we get

and Again put $(x+4) = u,$ Then $dx = du$ and changing Limit, we get

$\displaystyle \int_{-8}^{-10}\ln(t^2)dt+2\int_{4}^{5}\ln(u)^2du$

Now How can I solve after That

Help me

Thanks

Since both of these are related to $$\int (\ln(x))^2 \;dx$$ I will do this.

We use integration by parts with $$u=\ln(x)$$ and $$dv=\ln(x)$$. Then $$du=1/x$$ and $$v=x\ln(x)-x$$, and so:

$$\int (\ln(x))^2\;dx=\int u\; dv= u\; v - \int du\;v=x\ln(x)(\ln(x)-1) -\int (\ln(x) -1) \;dx$$

(note: I assume that $$x$$ is positive over any range of integration we may use, which can always be arranged as long as $$0$$ is not in the range)

(note: This is also a standard integral, item 15.530 in Scham's Handbook)

 

[Prove It]

Since both of these are related to $$\int (\ln(x))^2 \;dx$$ I will do this.

No they're not, they're related to $\displaystyle \begin{align*} \ln{ \left( x^2 \right) } \end{align*}$, not $\displaystyle \begin{align*} \left[ \ln{(x)} \right] ^2 \end{align*}$.