The given integral is:

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

To solve this, let's proceed step by step:

Step 1: Use Integration by Parts

The formula for integration by parts is: $\int u \, dv = uv - \int v \, du$ Here, we'll choose:

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

Now apply the formula: $\int \tan^{-1}(x) \, dx = x \tan^{-1}(x) - \int x \cdot \frac{1}{1+x^2} \, dx$

Step 2: Simplify the Remaining Integral

The remaining integral is: $\int \frac{x}{1+x^2} \, dx$ Let $w = 1 + x^2$, so $dw = 2x \, dx$. This substitution gives: $\int \frac{x}{1+x^2} \, dx = \frac{1}{2} \int \frac{1}{w} \, dw = \frac{1}{2} \ln|w| + C$ Substitute $w = 1 + x^2$ back: $\int \frac{x}{1+x^2} \, dx = \frac{1}{2} \ln(1+x^2) + C$

Step 3: Combine the Results

Now substitute this back into the integration by parts result: $\int \tan^{-1}(x) \, dx = x \tan^{-1}(x) - \frac{1}{2} \ln(1+x^2) + C$

Step 4: Apply the Definite Integral

Now evaluate the definite integral from 0 to 1: $\int_0^1 \tan^{-1}(x) \, dx = \left[ x \tan^{-1}(x) - \frac{1}{2} \ln(1+x^2) \right]_0^1$

At $x = 1$: $1 \cdot \tan^{-1}(1) - \frac{1}{2} \ln(1+1^2) = \frac{\pi}{4} - \frac{1}{2} \ln(2)$

At $x = 0$: $0 \cdot \tan^{-1}(0) - \frac{1}{2} \ln(1+0^2) = 0 - \frac{1}{2} \ln(1) = 0$

Subtract the two: $\int_0^1 \tan^{-1}(x) \, dx = \frac{\pi}{4} - \frac{1}{2} \ln(2)$

Final Answer:

$\boxed{\int_0^1 \tan^{-1}(x) \, dx = \frac{\pi}{4} - \frac{1}{2} \ln(2)}$

Do you need further clarification or additional examples?

Related Questions:

1. What are the properties of the arctangent function?
2. How is integration by parts derived and when is it used?
3. Can substitution methods always simplify integrals?
4. What is the geometric interpretation of the given integral?
5. How do logarithmic functions arise in integration problems?

Tip:

Always check the limits carefully in definite integrals when substitution is involved, as skipping this can lead to errors.