A-Level Techniques of integration: worked solution

Question

Using the previous parts, find the integral of x ln sqrt(x^2 + 1) / (x^2 + 1)^2 with respect to x.

The integrand is a logarithm multiplied by an algebraic factor, so use integration by parts, INT u dv = uv - INT v du. Choose u = ln sqrt(x² + 1), because a logarithm is easy to differentiate but awkward to integrate, and dv = x/(x² + 1)² dx, which is the piece the earlier part already integrated. Differentiate u: first use the log law ln sqrt(A) = (1/2) ln A, so u = (1/2) ln(x² + 1). Then du/dx = (1/2) * (2x)/(x² + 1) = x/(x² + 1). Integrate dv: for INT x (x² + 1)⁻² dx substitute w = x² + 1, so dw = 2x dx and x dx = (1/2) dw. INT x (x² + 1)⁻² dx = INT (1/2) w⁻² dw = (1/2)(-w⁻¹) = -1/(2(x² + 1)). So v = -1/(2(x² + 1)). Now apply the formula: INT x ln sqrt(x² + 1)/(x² + 1)² dx = [ln sqrt(x² + 1)] * [-1/(2(x² + 1))] - INT [-1/(2(x² + 1))] * [x/(x² + 1)] dx = -ln sqrt(x² + 1)/(2(x² + 1)) + (1/2) INT x/(x² + 1)² dx. The integral that is left is the same one done above, INT x/(x² + 1)² dx = -1/(2(x² + 1)). = -ln sqrt(x² + 1)/(2(x² + 1)) + (1/2) * (-1/(2(x² + 1))) + c = -ln sqrt(x² + 1)/(2(x² + 1)) - 1/(4(x² + 1)) + c. Check by differentiating. Writing the result as -(ln(x² + 1) + 1)/(4(x² + 1)), the quotient rule gives d/dx = -x(1 - ln(x² + 1))/(2(x² + 1)²) + x/(2(x² + 1)²) = x ln(x² + 1)/(2(x² + 1)²) = x ln sqrt(x² + 1)/(x² + 1)², which is the original integrand. So the integral is -ln sqrt(x² + 1)/(2(x² + 1)) - 1/(4(x² + 1)) + c.