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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

Step 1: Identify the Integral Structure

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Answer

To solve the integral ( \int \frac{1}{x^2 + 4x + 10} ; dx ), we need to complete the square in the denominator. The expression ( x^2 + 4x + 10 ) can be rewritten as:

[ x^2 + 4x + 10 = (x + 2)^2 + 6 ]

Step 2

Step 2: Substitute to Simplify the Integral

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Answer

Now we substitute ( u = x + 2 ), leading to:

[ du = dx ] Thus, the integral transforms into:

[ \int \frac{1}{u^2 + 6} ; du ]

Step 3

Step 3: Apply the Integral Formula

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Answer

The integral ( \int \frac{1}{u^2 + a^2} ; du = \frac{1}{a} \tan^{-1}\left( \frac{u}{a} \right) + c ) applies here with ( a = \sqrt{6} ):

[ \int \frac{1}{u^2 + 6} ; du = \frac{1}{\sqrt{6}} \tan^{-1}\left( \frac{u}{\sqrt{6}} \right) + c ]

Step 4

Step 4: Back Substitute and Select the Correct Expression

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Answer

Returning to original variable, we substitute back ( u = x + 2 ):

[ \frac{1}{\sqrt{6}} \tan^{-1}\left( \frac{x + 2}{\sqrt{6}} \right) + c ] This corresponds to option A in the original question.

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