Evaluate ∫ dx/(x^2 + 4x + 13).

Indefinite Integrals · Question 56 of 90

Question 56

HardOriginal question

Evaluate $\int \frac{dx}{x^2 + 4x + 13}$ .

\tan^{-1}\frac{x + 2}{3} + C

Correct answerYour answer

\frac{1}{3}\tan^{-1}\frac{x - 2}{3} + C

Correct answerYour answer

\frac{1}{3}\tan^{-1}\frac{x + 2}{3} + C

Correct answerYour answer

\log\left|x^2 + 4x + 13\right| + C

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

Complete the square: $x^2 + 4x + 13 = (x + 2)^2 + 9$ .
With $t = x + 2$ and $a = 3$ : $\int \frac{dt}{t^2 + 9} = \frac{1}{3}\tan^{-1}\frac{x + 2}{3} + C$ .

Worth noting

$\frac{x - 2}{3}$ gets the sign of the shift wrong: $x^2 + 4x$ is $(x + 2)^2 - 4$ .

Ten questions from Indefinite Integrals, at your own pace.

Ten questions from Indefinite Integrals, at your own pace.