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 (a)(iii) Linear combination or open all working together.

(b)(i)

Radical substitution

By using u=xu=\sqrt x, find ∫x1+x dx\int\dfrac{\sqrt x}{1+\sqrt x}\,dx.

Explore this idea

Hint

  1. x=u2x=u^2 so Differentiate x=u2x=u^2 to obtain dx=2u dudx=2u\,du. Equivalently, for x>0x>0, differentiate u=xu=\sqrt x and invert its derivative. Use u≥0u\ge0; then 1+u>01+u>0, which justifies the logarithm without modulus.

  2. Divide to get a polynomial plus a simple fraction.

Answer · (b)(i)

x−2x+2ln⁡(1+x)+Cx-2\sqrt x+2\ln(1+\sqrt x)+C.

x≥0x\ge0; derivative on x>0.

Explore the idea

Open interactive exploration

A curve rising towards 1

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 = 4.50
Reset the controls

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
2.534
Below-axis contribution
0
Signed integral
2.534

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

Signed integral from 0.00 to x
2.534
antiderivative − F(0)
2.534
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

For large xx the area grows almost linearly, because the integrand tends to 1.

Ideas to try
The integrand approaches 1 for large xx: what does that make the area?