Integration

9 methods · 36 questions

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The power rule only applies to terms $x^{k}$ . Before integrating, expand brackets and squares, split fractions term by term, and rewrite roots and reciprocals as fractional or negative powers so every term is integrable directly.

Trigger: an integrand that is a product, a square, a quotient, or contains $x$ or $1/ x^{n}$ .

Instances:
(i) split a single fraction over several numerator terms, e.g. $(3 - 2 x) / x^{3/2} = 3 x^{- 3/2} - 2 x^{- 1/2}$ ;
(ii) expand a square such as $(x^{2} - 4/ x^{2})^{2}$ into separate power terms;
(iii) cancel a factor first with a difference of two squares, e.g. $(t - 1) / (1 + t) = t - 1$ .

Linked questions (3)

TMUA 2016 Paper 2 · Q1

Find the value of

\int_{1}^{2} (x^{2} - \frac{4}{x ^{2}})^{2} d x

A $\frac{43}{15}$
B $3$
C $\frac{97}{15}$
D $\frac{103}{15}$
E $\frac{163}{15}$
F $18$

Worked solution

Answer:A

Expand the square:

(x^{2} - \frac{4}{x ^{2}})^{2} = x^{4} - 8 + 16 x^{- 4}

. Then integrate term by term:

\int_{1}^{2} (x^{4} - 8 + 16 x^{- 4}) d x = [\frac{x ^{5}}{5} - 8 x + \frac{16 x ^{- 3}}{- 3}]_{1}^{2} .

x = 2

\frac{32}{5} - 16 - \frac{16}{24} = \frac{32}{5} - 16 - \frac{2}{3}

. At

x = 1

\frac{1}{5} - 8 - \frac{16}{3}

. Difference:

(\frac{32}{5} - 16 - \frac{2}{3}) - (\frac{1}{5} - 8 - \frac{16}{3}) = \frac{31}{5} - 8 + \frac{14}{3} = \frac{93 - 120 + 70}{15} = \frac{43}{15}

. Hence the answer is A.

TMUA 2017 Paper 1 · Q1

Given that

\frac{d y}{d x} = 3 x^{2} - \frac{2 - 3 x}{x ^{3}}, x \neq = 0

and

y = 5

when

x = 1

, find

y

in terms of

x

A $y = \frac{1}{3} x^{3} + x^{- 2} - 3 x^{- 1} + 6 \frac{2}{3}$
B $y = x^{3} + \frac{1}{2} x^{- 2} - 3 x^{- 1} + 6 \frac{1}{2}$
C $y = x^{3} + x^{- 2} - 3 x^{- 1} + 6$
D $y = x^{3} + x^{- 2} - x^{- 1} + 4$
E $y = x^{3} + 2 x^{- 2} - x^{- 1} + 3$
F $y = 3 x^{3} + x^{- 2} - x^{- 1} + 2$

Worked solution

Answer:C

Rewrite the derivative by splitting the fraction:

\frac{d y}{d x} = 3 x^{2} - \frac{2}{x ^{3}} + \frac{3 x}{x ^{3}} = 3 x^{2} - 2 x^{- 3} + 3 x^{- 2} .

Integrate term by term:

y = x^{3} + x^{- 2} - 3 x^{- 1} + c .

Apply the initial condition

y = 5

when

x = 1

5 = 1 + 1 - 3 + c = - 1 + c ⟹ c = 6.

Hence

y = x^{3} + x^{- 2} - 3 x^{- 1} + 6

, which is option C.

TMUA 2018 Paper 1 · Q1

Find the value of

\int_{1}^{4} \frac{3 - 2 x}{x x} d x

A $- \frac{13}{2}$
B $- \frac{85}{16}$
C $- \frac{13}{8}$
D $- 1$
E $- \frac{1}{4}$
F $\frac{7}{4}$
G $7$

Worked solution

Answer:D

Rewrite the integrand by splitting the fraction and expressing in terms of powers of

x

\frac{3 - 2 x}{x x} = \frac{3 - 2 x}{x ^{3/2}} = 3 x^{- 3/2} - 2 x^{- 1/2} .

Integrate term by term:

\int_{1}^{4} (3 x^{- 3/2} - 2 x^{- 1/2}) d x = [\frac{3 x ^{- 1/2}}{- 1/2} - \frac{2 x ^{1/2}}{1/2}]_{1}^{4} = [- 6 x^{- 1/2} - 4 x^{1/2}]_{1}^{4} .

Evaluate: at

x = 4

- 6 \cdot \frac{1}{2} - 4 \cdot 2 = - 3 - 8 = - 11

. At

x = 1

- 6 \cdot 1 - 4 \cdot 1 = - 10

. So the integral equals

- 11 - (- 10) = - 1

. Hence the answer is D.