Answer to Question #101126 in Calculus for Jason

Question

Integrate both points with the placement method:

a) ∫ x√x+1 dx

b) ∫ 2x/x2+1 dx

place a at position t = x + 1 and at position b = t = x2 + 1

Answer 1

1) $\int x \sqrt{x+1} dx = [t = x+1; dt = dx; x = t-1] = \int (t-1)\sqrt{t} dt$.

The last expression contains only table integrals, hence $\int (t-1)\sqrt{t} dt = \frac{2}{5} t^{5/2} - \frac{2}{3}t^{3/2} + C = \frac{2}{5}(x+1)^{5/2}-\frac{2}{3}(x+1)^{3/2} + C$.

2) $\int 2 x \sqrt{x^2+1} dx = [t = x^2 + 1; dt = 2 x dx] = \int \sqrt{t} dt = \frac{2}{3}t^{3/2} + C$. Returning back to x variable, obtain $\int 2 x \sqrt{x^2+1} dx = \frac{2}{3}(x^2+1)^{3/2} + C$.