Which expression is equal to
\[ \int \frac{1}{x^2 + 4x + 10} \, dx \] ?
A - HSC - SSCE Mathematics Extension 2 - Question 6 - 2020 - Paper 1
Question 6
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
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
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
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
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
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
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.