Computing Indefinite Integrals — Question 8

Question 8 - Solution

Using substitution does not work! $$u = x^2 + 1, \qquad du = 2x\,dx.$$

Rewrite the integrand: $$(x^2+1)\ln(x^2+1)\,dx = u\ln u \cdot \frac{du}{2x}.$$

Instead, apply integration by parts directly.

Let $$u = \ln(x^2+1), \qquad dv = (x^2+1)\,dx.$$ Then $$du = \frac{2x}{x^2+1}\,dx, \qquad v = \frac{x^3}{3} + x.$$

Apply integration by parts: $$\int (x^2+1)\ln(x^2+1)\,dx = \left(\frac{x^3}{3}+x\right)\ln(x^2+1) - \int \left(\frac{x^3}{3}+x\right)\frac{2x}{x^2+1}\,dx.$$

Simplify the remaining integral: $$\left(\frac{x^3}{3}+x\right)\frac{2x}{x^2+1} = \frac{2x^4}{3(x^2+1)} + \frac{2x^2}{x^2+1}.$$

Use algebraic division: $$\frac{x^4}{x^2+1} = x^2 - 1 + \frac{1}{x^2+1}.$$

Thus, $$\int (x^2+1)\ln(x^2+1)\,dx = \left(\frac{x^3}{3}+x\right)\ln(x^2+1) - \int \left(\frac{2}{3}(x^2-1) + \frac{2}{3(x^2+1)} + \frac{2x^2}{x^2+1}\right)dx.$$

Compute term by term: $$\int (x^2-1)\,dx = \frac{x^3}{3} - x,$$ $$\int \frac{1}{x^2+1}\,dx = \arctan x, \qquad \int \frac{x^2}{x^2+1}\,dx = x - \arctan x.$$

Putting everything together, $$\boxed{ \int (x^2+1)\ln(x^2+1)\,dx = \left(\frac{x^3}{3}+x\right)\ln(x^2+1) - \frac{2x^3}{9} - \frac{4x}{3} + \frac{4}{3}\arctan x + C }$$