How can we solve this integral: ∫xdx/ (1+x^2) by integration by parts? What is its solution, if possible?

To solve the integral ∫(x dx) / (1 + x^2) using integration by parts, you can apply the following formula for integration by parts:

∫u dv = uv – ∫v du

In this case, we’ll choose the following assignments:

u = x (Differentiable part) dv = dx / (1 + x^2) (Integrable part)

Now, let’s calculate the derivatives and integrals of u and dv:

du = dx v = ∫dx / (1 + x^2)

Now, we need to find v, which involves integrating ∫dx / (1 + x^2). This integral can be solved using a trigonometric substitution.

Let’s substitute x with tan(θ) and dx with sec^2(θ) dθ:

∫dx / (1 + x^2) = ∫sec^2(θ) dθ

This is a standard integral, and its antiderivative is simply tan(θ) + C, where C is the constant of integration.

Now, we need to go back to the original variable x. Recall that x = tan(θ):

v = tan(θ) + C

Now, we can apply the integration by parts formula:

∫(x dx) / (1 + x^2) = uv – ∫v du ∫(x dx) / (1 + x^2) = x(tan(θ) + C) – ∫(tan(θ) + C) dx

Now, we can integrate the remaining expression with respect to θ:

∫(tan(θ) + C) dx = ∫tan(θ) dx + ∫C dx

Integrating tan(θ) with respect to θ gives us -ln|cos(θ)| + C1, where C1 is another constant of integration.

So, our integral becomes:

∫(x dx) / (1 + x^2) = x(tan(θ) + C) + ln|cos(θ)| + C1

Now, remember that x = tan(θ), so we need to find θ in terms of x. We know that tan(θ) = x, so θ = arctan(x).

Putting it all together:

∫(x dx) / (1 + x^2) = x(arctan(x) + C) + ln|cos(arctan(x))| + C1

Now, we need to simplify the expression further. Since cos(arctan(x)) = 1 / √(1 + x^2), we have:

∫(x dx) / (1 + x^2) = x(arctan(x) + C) + ln|1 / √(1 + x^2)| + C1

So, that’s the solution to the integral ∫(x dx) / (1 + x^2) using integration by parts.