∫(x+3)√(x^2+2x+5) dx is equal to (C is a constant of integration)

The correct answer is Option (4) → \(\displaystyle\frac{1}{3}\left(x^2+2x+5\right)^{3/2}+(x+1)\sqrt{x^2+2x+5}+4\log\left|x+1+\sqrt{x^2+2x+5}\right|+C\)

The derivative of \(x^2+2x+5\) is 2x + 2, so the linear factor is written as a multiple of the derivative plus a constant:

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

So

    \( \displaystyle I = \frac{1}{2}\int(2x+2)\sqrt{x^2+2x+5}\,dx+2\int\sqrt{x^2+2x+5}\,dx \)

In the first part put \(t = x^2+2x+5\), \(dt = (2x+2)\,dx\):

    \( \displaystyle \frac{1}{2}\int t^{1/2}\,dt = \frac{1}{3}\left(x^2+2x+5\right)^{3/2} \)

In the second part complete the square, \(x^2+2x+5 = (x+1)^2+2^2\), and use the standard result \(\int\sqrt{u^2+a^2}\,du = \frac{u}{2}\sqrt{u^2+a^2}+\frac{a^2}{2}\log\left|u+\sqrt{u^2+a^2}\right|\):

    \( \displaystyle 2\int\sqrt{(x+1)^2+2^2}\,dx = 2\left[\frac{x+1}{2}\sqrt{x^2+2x+5}+2\log\left|x+1+\sqrt{x^2+2x+5}\right|\right] \)

    \( \displaystyle = (x+1)\sqrt{x^2+2x+5}+4\log\left|x+1+\sqrt{x^2+2x+5}\right| \)

Adding the two parts:

    \( \displaystyle I = \frac{1}{3}\left(x^2+2x+5\right)^{3/2}+(x+1)\sqrt{x^2+2x+5}+4\log\left|x+1+\sqrt{x^2+2x+5}\right|+C \)

Why the other options are incorrect

• Option (1) \(\displaystyle\frac{1}{3}\left(x^2+2x+5\right)^{3/2}+(x+1)\sqrt{x^2+2x+5}+2\log\left|x+1+\sqrt{x^2+2x+5}\right|+C\) — Incorrect. It forgets to multiply the log term of the standard result by the outside factor 2.

• Option (2) \(\displaystyle\frac{2}{3}\left(x^2+2x+5\right)^{3/2}+(x+1)\sqrt{x^2+2x+5}+4\log\left|x+1+\sqrt{x^2+2x+5}\right|+C\) — Incorrect. It leaves out the factor \(\frac{1}{2}\) in front of \(\int(2x+2)\sqrt{x^2+2x+5}\,dx\).

• Option (3) \(\displaystyle\frac{1}{3}\left(x^2+2x+5\right)^{3/2}+2(x+1)\sqrt{x^2+2x+5}+4\log\left|x+1+\sqrt{x^2+2x+5}\right|+C\) — Incorrect. The factor \(\frac{1}{2}\) of the standard result cancels the outside 2, leaving \((x+1)\sqrt{x^2+2x+5}\).