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Which expression is equal to \[ \int \frac{1}{x^2 + 4x + 10} \, dx \] ? A - HSC - SSCE Mathematics Extension 2 - Question 6 - 2020 - Paper 1

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Which-expression-is-equal-to--\[-\int-\frac{1}{x^2-+-4x-+-10}-\,-dx-\]-?--A-HSC-SSCE Mathematics Extension 2-Question 6-2020-Paper 1.png

Which expression is equal to \[ \int \frac{1}{x^2 + 4x + 10} \, dx \] ? A. \( \frac{1}{\sqrt{6}} \tan^{-1}\left( \frac{x + 2}{\sqrt{6}} \right) + c \) B. \( \tan^... show full transcript

Worked Solution & Example Answer:Which expression is equal to \[ \int \frac{1}{x^2 + 4x + 10} \, dx \] ? A - HSC - SSCE Mathematics Extension 2 - Question 6 - 2020 - Paper 1

Step 1

Identify the integral

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Answer

The integral we need to evaluate is [ \int \frac{1}{x^2 + 4x + 10} , dx. ] First, we can complete the square in the denominator. The quadratic expression can be rewritten as: [ x^2 + 4x + 10 = (x + 2)^2 + 6. ]

Step 2

Use substitution for integration

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Answer

Next, we'll perform a substitution. Let: [ u = x + 2, ] in which case, ( dx = du ), and the integral becomes: [ \int \frac{1}{u^2 + 6} , du. ] This integral can be solved using the formula: [ \int \frac{1}{a^2 + u^2} , du = \frac{1}{a} \tan^{-1}\left( \frac{u}{a} \right) + c. ] Here, ( a = \sqrt{6} ). So the integral simplifies to: [ \frac{1}{\sqrt{6}} \tan^{-1}\left( \frac{u}{\sqrt{6}} \right) + c. ]

Step 3

Back substitute to original variable

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Answer

Replacing ( u ) with ( x + 2 ), we have: [ \frac{1}{\sqrt{6}} \tan^{-1}\left( \frac{x + 2}{\sqrt{6}} \right) + c. ] Thus, the answer corresponds to option A.

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