Evaluate ∫ tan⁻¹(√x) dx.

To evaluate the integral ∫ tan⁻¹(√x) dx, we can use integration by parts. Let's start by setting:

Choosing u and dv

Let:

• u = tan⁻¹(√x) (which we will differentiate)
• dv = dx (which we will integrate)

Finding du and v

Now, we need to find du and v:

• To find du, we differentiate u:

Using the derivative of tan⁻¹(x), we have:

du = (1 / (1 + (√x)²)) * (1 / (2√x)) dx = (1 / (2x + 2)) dx

• To find v, we integrate dv:

v = ∫dx = x

Applying Integration by Parts

Using the integration by parts formula, ∫u dv = uv - ∫v du, we get:

∫ tan⁻¹(√x) dx = x * tan⁻¹(√x) - ∫ x * (1 / (2x + 2)) dx

Simplifying the Remaining Integral

The remaining integral can be simplified:

∫ x / (2x + 2) dx = (1/2) ∫ (x / (x + 1)) dx

This can be further simplified using long division:

∫ (1 - 1 / (x + 1)) dx = ∫ dx - ∫ (1 / (x + 1)) dx

Final Steps

Now we can integrate:

• ∫ dx = x
• ∫ (1 / (x + 1)) dx = ln|x + 1|

Putting It All Together

Combining everything, we have:

∫ tan⁻¹(√x) dx = x * tan⁻¹(√x) - (1/2)(x - ln|x + 1|) + C

Thus, the final result is:

∫ tan⁻¹(√x) dx = x * tan⁻¹(√x) - (1/2)x + (1/2)ln|x + 1| + C