12 min readPathway sheet2-mark and 3-mark questions; the (px + q) type is a common 3-mark answer

Integrals with ax² + bx + c in the denominator

Complete the square in the quadratic, then use a special integral. If px + q is on top, first split it using the quadratic's derivative.

Completing the squarex² + bx + c = (x + b/2)² + (c − b²/4). If the coefficient of x² is not 1, take it out first.

Split for px + qpx + q = A × d/dx (ax² + bx + c) + B = A(2ax + b) + B. Compare coefficients to get A and B.

Two quick results∫ (2ax + b)/(ax² + bx + c) dx = log|ax² + bx + c| + C; ∫ (2ax + b)/√(ax² + bx + c) dx = 2√(ax² + bx + c) + C

Method A: dx/(ax² + bx + c) and dx/√(ax² + bx + c)

1
Take the coefficient of x² outside. Under a root, take out its square root.
2
Add and subtract (half the coefficient of x)².
3
Write the quadratic as (x + k)² + m², (x + k)² − m² or m² − (x + k)².
4
Put t = x + k, so dt = dx.
5
Use the matching special integral, then put back t = x + k.

Method B: (px + q) on top

1
Write px + q = A(2ax + b) + B, where 2ax + b is the derivative of the quadratic.
2
Compare the coefficients of x, then the constant terms, to find A and B.
3
Split: A ∫ (2ax + b)/(ax² + bx + c) dx + B ∫ dx/(ax² + bx + c).
4
The first part is A log|ax² + bx + c|. Under a root, it is 2A√(ax² + bx + c).
5
Do the second part by Method A. Add one C at the end.

Example 1: tan⁻¹ type

Problem

Find ∫ dx/(2x² + 8x + 26).

Working

Take 2 out: 2x² + 8x + 26 = 2(x² + 4x + 13).
Complete the square: x² + 4x + 13 = (x + 2)² − 4 + 13 = (x + 2)² + 3².
The integral is (1/2) ∫ dx/((x + 2)² + 3²).
Put t = x + 2, dt = dx, and use ∫ dt/(t² + a²) = (1/a) tan⁻¹(t/a) + C with a = 3.
= (1/2)(1/3) tan⁻¹((x + 2)/3) + C.

Answer

(1/6) tan⁻¹((x + 2)/3) + C

Example 2: sin⁻¹ type

Problem

Find ∫ dx/√(8 + 2x − x²).

Working

Take −1 out of the x terms: 8 + 2x − x² = 8 − (x² − 2x).
Complete the square: x² − 2x = (x − 1)² − 1.
So 8 + 2x − x² = 8 − (x − 1)² + 1 = 3² − (x − 1)².
Put t = x − 1 and use ∫ dt/√(a² − t²) = sin⁻¹(t/a) + C with a = 3.

Answer

sin⁻¹((x − 1)/3) + C

Example 3: (px + q) over a quadratic

Problem

Find ∫ (3x + 1)/(x² + 2x + 5) dx.

Working

The derivative of x² + 2x + 5 is 2x + 2. Write 3x + 1 = A(2x + 2) + B.
Coefficient of x: 2A = 3, so A = 3/2. Constants: 2A + B = 1, so B = −2.
Integral = (3/2) ∫ (2x + 2)/(x² + 2x + 5) dx − 2 ∫ dx/(x² + 2x + 5).
First part = (3/2) log(x² + 2x + 5). No modulus is needed, as x² + 2x + 5 = (x + 1)² + 4 is always positive.
Second part: −2 ∫ dx/((x + 1)² + 2²) = −2 × (1/2) tan⁻¹((x + 1)/2) = −tan⁻¹((x + 1)/2).

Answer

(3/2) log(x² + 2x + 5) − tan⁻¹((x + 1)/2) + C

Example 4: (px + q) over a root (board style)

Problem

Find ∫ (x + 2)/√(8 + 2x − x²) dx.

Working

The derivative of 8 + 2x − x² is 2 − 2x. Write x + 2 = A(2 − 2x) + B.
Coefficient of x: −2A = 1, so A = −1/2. Constants: 2A + B = 2, so B = 3.
Integral = −(1/2) ∫ (2 − 2x)/√(8 + 2x − x²) dx + 3 ∫ dx/√(8 + 2x − x²).
First part: −(1/2) × 2√(8 + 2x − x²) = −√(8 + 2x − x²).
Second part, from Example 2: 3 sin⁻¹((x − 1)/3).

Answer

−√(8 + 2x − x²) + 3 sin⁻¹((x − 1)/3) + C

In real life

Completing the square turns a messy quadratic into a clean square plus a constant. The same trick finds the lowest point of a cost curve in a factory or the highest point of a cricket ball's path.

Exam angle

2 marks: “Find ∫ dx/(x² + 4x + 8).” x² + 4x + 8 = (x + 2)² + 2². Use the tan⁻¹ form with a = 2: (1/2) tan⁻¹((x + 2)/2) + C.

3 marks: “Find ∫ (2x − 1)/(x² + x + 1) dx.” Write 2x − 1 = (2x + 1) − 2. First part: log(x² + x + 1). Second part: −2 ∫ dx/((x + 1/2)² + (√3/2)²) = −(4/√3) tan⁻¹((2x + 1)/√3). Answer: log(x² + x + 1) − (4/√3) tan⁻¹((2x + 1)/√3) + C.

3 marks: “Find ∫ x/√(x² + 4x + 13) dx.” Write x = (1/2)(2x + 4) − 2. First part: √(x² + 4x + 13). Second part: −2 ∫ dx/√((x + 2)² + 3²) = −2 log|x + 2 + √(x² + 4x + 13)|. Answer: √(x² + 4x + 13) − 2 log|x + 2 + √(x² + 4x + 13)| + C.

Remember it

Square the bottom, split the top. Complete the square below; write the top as A × (derivative of the bottom) + B.