Evaluate (2x+3)¹/2} δ x

Correct Answer is : 13(2x+3)32+k1/3 (2x+3)³/2} + k31(2x+3)23+k

Correct Answer is : 13(2x+3)32+k1/3 (2x+3)³/2} + k31(2x+3)23+k

Test Index

JAMB Mathematics 2014 Paper

Question : 33

Total: 47

Evaluate

\int (2x+3)^{\frac{1}{2}} \delta x

Solution:

Concept:

Use the substitution method to integrate a composite function by reversing the chain rule.

Explanation:

Let

u = 2x + 3

Then

\frac{\mathrm{d}u}{\mathrm{d}x} = 2

, so

\mathrm{d}x = \frac{\mathrm{d}u}{2}

Substitute into the integral:

\int (2x+3)^{\frac{1}{2}} \, \mathrm{d}x = \int u^{\frac{1}{2}} \cdot \frac{\mathrm{d}u}{2}

Factor out the constant:

= \frac{1}{2} \int u^{\frac{1}{2}} \, \mathrm{d}u

Apply the power rule for integration:

\frac{1}{2} \cdot \frac{u^{\frac{3}{2}}}{\frac{3}{2}} + k = \frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}} + k

Simplify:

= \frac{1}{3} u^{\frac{3}{2}} + k

Replace

u

with

2x + 3

= \frac{1}{3} (2x+3)^{\frac{3}{2}} + k

Answer:

\frac{1}{3} (2x+3)^{\frac{3}{2}} + k

This corresponds to option C.

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