# Definite Integral Challenge

#### anemone

##### MHB POTW Director
Staff member
Evaluate $$\displaystyle \int_2^4 \frac{\sqrt{\ln (9-x)}\,dx}{\sqrt{\ln (9-x)}+\sqrt{\ln (x+3)}}$$

#### MarkFL

Staff member
My solution:

We are given to evaluate:

$$\displaystyle I=\int_2^4 \frac{\sqrt{\ln (9-x)}\,dx}{\sqrt{\ln (9-x)}+\sqrt{\ln (x+3)}}$$

Let:

$$\displaystyle u=x-3\,\therefore\,du=dx$$

and we have:

$$\displaystyle I=\int_{-1}^{1}\frac{\sqrt{\ln(6-u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}\,du$$

Using the property of definite integrals:

$$\displaystyle \int_{-a}^a f(x)\,dx=\int_0^a \left[f(x)+f(-x) \right]\,dx$$

we may write:

$$\displaystyle I=\int_{0}^{1}\frac{\sqrt{\ln(6-u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}+\frac{\sqrt{\ln(6+u)}}{\sqrt{\ln(6+u)}+\sqrt{\ln(6-u)}}\,du$$

$$\displaystyle I=\int_{0}^{1}\frac{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}\,du$$

$$\displaystyle I=\int_{0}^{1}\,du=_0^1=1-0=1$$

#### anemone

##### MHB POTW Director
Staff member
My solution:

We are given to evaluate:

$$\displaystyle I=\int_2^4 \frac{\sqrt{\ln (9-x)}\,dx}{\sqrt{\ln (9-x)}+\sqrt{\ln (x+3)}}$$

Let:

$$\displaystyle u=x-3\,\therefore\,du=dx$$

and we have:

$$\displaystyle I=\int_{-1}^{1}\frac{\sqrt{\ln(6-u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}\,du$$

Using the property of definite integrals:

$$\displaystyle \int_{-a}^a f(x)\,dx=\int_0^a \left[f(x)+f(-x) \right]\,dx$$

we may write:

$$\displaystyle I=\int_{0}^{1}\frac{\sqrt{\ln(6-u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}+\frac{\sqrt{\ln(6+u)}}{\sqrt{\ln(6+u)}+\sqrt{\ln(6-u)}}\,du$$

$$\displaystyle I=\int_{0}^{1}\frac{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}\,du$$

$$\displaystyle I=\int_{0}^{1}\,du=_0^1=1-0=1$$

That's a smart approach, MarkFL! Well done!

#### Jester

##### Well-known member
MHB Math Helper
A little more direct
Let $x = 6-u$ so the integral become

$I = \displaystyle - \int_4^2 \dfrac{\sqrt{\ln (3+u)}}{ \sqrt{\ln (3+u)} +\sqrt{\ln (9-u)}}\;du$

Replace $u$ with $x$ and add to the original and like Mark said $2I = \int_2^4 1dx$ gives $I = 1$