Reuse two basic integrals and a radical substitution

(a)(i) Find ∫2−x4x−x2−1dx\int\frac{2-x}{\sqrt{4x-x^2-1}}dx. marks

(ii) Find ∫14x−x2−1dx\int\frac1{\sqrt{4x-x^2-1}}dx. marks

(iii) Using (i) and (ii), find ∫x4x−x2−1dx\int\frac x{\sqrt{4x-x^2-1}}dx. marks

(b)(i) By using u=xu=\sqrt x, find ∫x1+xdx\int\frac{\sqrt x}{1+\sqrt x}dx. marks

(ii) Find the exact value of ∫09ln⁡(1+x)dx\int_0^9\ln(1+\sqrt x)dx, leaving the answer aln⁡b−c/ba\ln b-c/b, where a,b,ca,b,c are integers to be determined. marks

[2016/VJC/Promo/11]

Total marks: 14

Solution

If this part uses an earlier result, revisit (b)(i) Radical substitution or open all working together.

(b)(ii)

Integration by parts with a given integral

Find the exact value of ∫09ln⁡(1+x) dx\int_0^9\ln(1+\sqrt x)\,dx, in the form aln⁡b−cba\ln b-\dfrac cb.

Explore this idea

Hint

  1. Parts with u=ln⁡(1+x)u=\ln(1+\sqrt x) and dv=dxdv=dx.

  2. The remaining integral is part (b)(i).

Answer · (b)(ii)

16ln⁡2−3/216\ln2-3/2; a=16,b=2,c=3a=16,b=2,c=3.

Explore the idea

Open interactive exploration

The area from 0 to 9

Drag x. The shading shows the selected interval; the lower curves compare the numerical signed integral with the change in the antiderivative.

Upper limit xx = 9.00
Try an example

How the shaded areas add up

The limits run left to right. Area above the axis adds; area below the axis subtracts.

Above-axis contribution
9.59
Below-axis contribution
0
Signed integral
9.59

All three bars share one scale. Positive contributions extend right of zero; negative contributions extend left.

Signed integral from 0.00 to x
9.59
antiderivative − F(0)
9.59
agrees at displayed precision
Formula − integral
0
0.00% of the integral’s magnitude
Compare the running integral on a graph
  • Integrand
  • Numerical signed integral from the start to x
  • Part of the integrand above the axis
  • Part of the integrand below the axis
  • F(x) − F(start), from the antiderivative
Key idea

The exact area is 16ln⁡2−3216\ln2-\frac32.

Ideas to try
Drag xx to 9: the readout is 16ln⁡2−32≈9.5916\ln2-\frac32\approx9.59.