Integration by Substitution

Edexcel IAL June 2021 P4 (WMA14/01) Q4 Integration by Substitution

Use algebraic integration and the substitution \( u = \sqrt{x} \) to find the exact value of

\[\int_{1}^{4} \frac{10}{5x + 2x\sqrt{x}} dx\]

Write your answer in the form \( 4\ln\left(\frac{a}{b}\right) \), where \( a \) and \( b \) are integers to be found.

(Solutions relying entirely on calculator technology are not acceptable.)

Solution: Setting Up the Integral and Evaluating the Integral

Key Concepts Used:

Algebraic Substitution: Using \( u = \sqrt{x} \) to simplify the integrand.
Differential Transformation: Expressing \( dx \) in terms of \( du \).
Limit Adjustment: Changing integration limits to match the new variable.
Partial Fractions: Decomposing complex rational expressions.
Logarithmic Integration: Integrating terms of the form \( \frac{1}{u} \).
Exact Value Evaluation: Calculating definite integral values.

Step-by-Step Solution:

1. Apply the Substitution

Given substitution:

\[u = \sqrt{x} = x^{1/2}\]

Compute differential:

\[\frac{du}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\]

\[dx = 2\sqrt{x} du = 2u du\]

2. Rewrite the Integrand

Original integrand: \(\frac{10}{5x + 2x\sqrt{x}}\)

Substitute \( x = u^2 \) and \( dx = 2u du \):

\[\frac{10}{5u^2 + 2u^3} \times 2u du = \frac{20u}{u^2 (5 + 2u)} du\]

Simplify:

\[\frac{20}{u(5 + 2u)} du\]

3. Change the Limits

Original limits \( x = 1 \) to \( x = 4 \):

When \( x = 1 \): \( u = 1 \)

When \( x = 4 \): \( u = 2 \)

4. Partial Fraction Decomposition

Express:

\[\frac{20}{u(5 + 2u)} = \frac{A}{u} + \frac{B}{5 + 2u}\]

Multiply through by denominator:

\[20 = A(5 + 2u) + B(u)\]

Solve for coefficients:

When \( u = 0 \): \( 20 = 5A \Rightarrow A = 4 \)

When \( u = -2.5 \): \( 20 = -2.5B \Rightarrow B = -8 \)

Thus:

\[\frac{20}{u(5 + 2u)} = \frac{4}{u} – \frac{8}{5 + 2u}\]

5. Integrate Term by Term

\[\int \left( \frac{4}{u} – \frac{8}{5 + 2u} \right) du = 4\ln|u| – 4\ln|5 + 2u| + C\]

6. Evaluate at Limits

At \( u = 2 \): \( 4\ln2 – 4\ln9 \)

At \( u = 1 \): \( 4\ln1 – 4\ln7 = 0 – 4\ln7 \)

Subtract lower limit from upper limit:

\[(4\ln2 – 4\ln9) – (-4\ln7)\]

\[= 4\ln2 – 4\ln9 + 4\ln7\]

7. Simplify the Expression

Factor out 4:

\[4(\ln2 – \ln9 + \ln7)\]

Combine logarithms:

\[4\ln\left( \frac{2 \times 7}{9} \right) = 4\ln\left( \frac{14}{9} \right)\]

Final Answer:

\[4\ln\left( \frac{14}{9} \right)\]

Coach Name: Sir Muhammad Abdullah Shah

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