The integral provided in the image is:

$\int_0^1 \tan^{-1}(x) \, dx$

Solution:

We need to evaluate this definite integral. The function $\tan^{-1}(x)$ (arctangent of $x$) does not have an elementary antiderivative when integrated with respect to $x$. However, we can solve it using integration by parts.

Step 1: Use integration by parts

The formula for integration by parts is: $\int u \, dv = uv - \int v \, du$

Here, let:

• $u = \tan^{-1}(x)$, so that $du = \frac{1}{1+x^2} \, dx$,
• $dv = dx$, so that $v = x$.

Step 2: Apply the formula

$\int_0^1 \tan^{-1}(x) \, dx = \left[ x \tan^{-1}(x) \right]_0^1 - \int_0^1 x \cdot \frac{1}{1+x^2} \, dx$

Step 3: Evaluate the first term

$\left[ x \tan^{-1}(x) \right]_0^1 = \left( 1 \cdot \tan^{-1}(1) \right) - \left( 0 \cdot \tan^{-1}(0) \right)$ $= \tan^{-1}(1) - 0 = \frac{\pi}{4}$

Step 4: Simplify the second integral

Now, evaluate: $\int_0^1 \frac{x}{1+x^2} \, dx$

Let $u = 1+x^2$, so that $du = 2x \, dx$. Then: $\int \frac{x}{1+x^2} \, dx = \frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln|u| + C = \frac{1}{2} \ln(1+x^2) + C$

Evaluate this from 0 to 1: [ \int_0^1 \frac{x}{1+x^2} , dx = \frac{1}{2} \left[ \ln(1+1^2) - \ln(1+0^2) \right] = \frac{1}{2} \left[ \ln(2) - \ln(1) \right] ] $= \frac{1}{2} \ln(2)$

Step 5: Combine the results

Substitute back into the formula: $\int_0^1 \tan^{-1}(x) \, dx = \frac{\pi}{4} - \frac{1}{2} \ln(2)$

Final Answer:

$\boxed{\frac{\pi}{4} - \frac{\ln(2)}{2}}$

Would you like further clarification or more examples? Here are five related questions to deepen your understanding:

1. How do you apply integration by parts in general?
2. Can you evaluate $\int \tan^{-1}(x) \, dx$ without definite limits?
3. What is the geometric interpretation of $\tan^{-1}(x)$?
4. How does substitution simplify the integral $\int \frac{x}{1+x^2} \, dx$?
5. What is the significance of the natural logarithm $\ln(x)$ in integration?

Tip: Integration by parts often simplifies integrals involving products of functions—choose $u$ and $dv$ wisely!