I see you've uploaded an image of two integrals that need to be solved. I'll provide solutions for both. Let’s go step by step:

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(الف)

$\int_0^1 \frac{\ln(x)}{\sqrt{1 - x}} \, dx$

Solution:

We will solve this using substitution:

1. Let $x = \sin^2(t)$, so $dx = 2\sin(t)\cos(t) \, dt = \sin(2t) \, dt$.
2. The limits transform as follows:
   • When $x = 0$, $t = 0$,
   • When $x = 1$, $t = \frac{\pi}{2}$.
3. Substitute into the integral: $\int_0^1 \frac{\ln(x)}{\sqrt{1 - x}} \, dx = \int_0^{\frac{\pi}{2}} \frac{\ln(\sin^2(t))}{\cos(t)} \cdot \sin(2t) \, dt.$
4. Simplify using $\ln(\sin^2(t)) = 2\ln(\sin(t))$ and $\sin(2t) = 2\sin(t)\cos(t)$: $\int_0^{\frac{\pi}{2}} \frac{\ln(\sin^2(t))}{\cos(t)} \cdot \sin(2t) \, dt = 2\int_0^{\frac{\pi}{2}} \ln(\sin(t)) \cdot 2\sin(t) \, dt.$ This reduces to: $4\int_0^{\frac{\pi}{2}} \ln(\sin(t)) \, dt.$

Result:

A known result is that: $\int_0^{\frac{\pi}{2}} \ln(\sin(t)) \, dt = -\frac{\pi}{2} \ln(2).$ Thus: $\int_0^1 \frac{\ln(x)}{\sqrt{1 - x}} \, dx = 4 \left(-\frac{\pi}{2} \ln(2)\right) = -2\pi \ln(2).$

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(ب)

$\int_1^e \frac{\ln(x)}{x \left(1 + (\ln(x))^2\right)} \, dx$

Solution:

We solve using substitution:

1. Let $u = \ln(x)$, so $du = \frac{1}{x} \, dx$.
2. The limits transform as follows:
   • When $x = 1$, $u = \ln(1) = 0$,
   • When $x = e$, $u = \ln(e) = 1$.
3. Substitute into the integral: $\int_1^e \frac{\ln(x)}{x \left(1 + (\ln(x))^2\right)} \, dx = \int_0^1 \frac{u}{1 + u^2} \, du.$

Simplify:

The integral becomes: $\int_0^1 \frac{u}{1 + u^2} \, du.$ Using substitution $v = 1 + u^2$, $dv = 2u \, du$, the limits transform as:

• When $u = 0$, $v = 1$,
• When $u = 1$, $v = 2$.

The integral becomes: $\int_0^1 \frac{u}{1 + u^2} \, du = \frac{1}{2} \int_1^2 \frac{1}{v} \, dv.$

Solve:

$\frac{1}{2} \int_1^2 \frac{1}{v} \, dv = \frac{1}{2} \ln(v) \bigg|_1^2 = \frac{1}{2} \left(\ln(2) - \ln(1)\right).$ Since $\ln(1) = 0$, the result is: $\frac{1}{2} \ln(2).$

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Final Answers:

1. (الف): $\int_0^1 \frac{\ln(x)}{\sqrt{1 - x}} \, dx = -2\pi \ln(2).$
2. (ب): $\int_1^e \frac{\ln(x)}{x \left(1 + (\ln(x))^2\right)} \, dx = \frac{1}{2} \ln(2).$

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Would you like further clarification on any step or a more detailed explanation?

Related Questions:

1. How can substitution techniques simplify challenging integrals?
2. What are some common integral results involving logarithmic functions?
3. How is symmetry used in definite integrals?
4. Can we solve the first integral without substitution?
5. Why is $\int_0^{\frac{\pi}{2}} \ln(\sin(t)) \, dt = -\frac{\pi}{2} \ln(2)$ a standard result?

Tip:

When solving integrals, recognize patterns that align with standard results to save time and effort.