Substitution to get to a specific form

Question

So I need to get the integral $$ \int_0^1 \frac{x+3}{x^2+x+2}dx $$ to the form $$ \int_a^b \frac{Au+B}{u^2+1}du $$ for some $a, b, A, B \in \mathbb R$. I can get either the numerator to the correct form, with the denominator being $u^2+\frac{7}{4}$, or the denominator to the correct form but not the numerator. Kind of stuck here...

Answer 1

If $x = su + t$ for constants $s, t$, then $$\begin{align} x^2 + x + 2 &= (su + t)^2 + (su + t) + 2 \\ &= s^2 u^2 + 2stu + t^2 + su + t + 2 \\ &= s^2 u^2 + (2st + s)u + (t^2 + t + 2). \end{align}$$ So we require $2st + s = 0$, or $t = -\frac{1}{2}$. Consequently, $t^2 + t + 2 = \frac{7}{4}$. Finally, we require $$s^2 = t^2 + t + 2 = \frac{7}{4},$$ hence $$s = \frac{\sqrt{7}}{2}.$$ It follows that the substitution $$x = \frac{\sqrt{7}}{2} u - \frac{1}{2}, \quad dx = \frac{\sqrt{7}}{2} \, du$$ or $$u = \frac{2x + 1}{\sqrt{7}}, \quad du = \frac{2}{\sqrt{7}} \, dx$$ is the desired transformation. I leave the determination of $a, b, A, B$ as an exercise.

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You may be wondering why we set $s^2 = t^2 + t + 2$. The reason is because after the substitution $x = su + t$, we could regard the result in the form $$s^2 \left(u^2 + \frac{2t+1}{s} u + \frac{t^2 + t + 2}{s^2}\right),$$ and the constant term $\frac{t^2 + t + 2}{s^2}$ in the monic quadratic factor must equal $1$.

Answer 2

Note: The other answer is great, but cosmetically dissimilar to mine, so I'll post this as well.

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1. Complete the square in the denominator. $$ x^2 + x + 2 = \bigl(x + \tfrac12 \bigr)^2 + \tfrac74 $$

2. Scale both numerator and denominator by constant from the denominator. $$ \frac{x+3}{\bigl(x + \tfrac12 \bigr)^2 + \tfrac74} = \frac{\tfrac47\bigl(x+3\bigr)}{\tfrac47 \bigl(x + \tfrac12 \bigr)^2 + 1} $$

3. Incorporate the reciprocal constant into the square. $$ \frac{\tfrac47\bigl(x+3\bigr)}{\tfrac47 \bigl(x + \tfrac12 \bigr)^2 + 1} = \frac47 \cdot \frac{x+3}{\Bigl(\tfrac{2}{\sqrt{7}} \bigl(x + \tfrac12 \bigr)\Bigr)^2 + 1} $$

4. Make substitution. Let $u = \tfrac{2}{\sqrt{7}} \bigl(x + \tfrac12 \bigr)$, so $x = \tfrac{\sqrt{7}}{2}u - \tfrac12$ and $x + 3 = \tfrac{\sqrt{7}}{2}u + \tfrac52$, and $$ \frac47 \cdot \frac{x+3}{\Bigl(\tfrac{2}{\sqrt{7}} \bigl(x + \tfrac12 \bigr)\Bigr)^2 + 1} = \frac47 \cdot \frac{\tfrac{\sqrt{7}}{2}u + \tfrac52}{u^2 + 1} $$

5. Make substitution with differentials as well. Finally some calculus! Since $\mathrm{d}x = \frac{\sqrt{7}}{2}\,\mathrm{d}u$, $$ \int_0^1 \frac{x+3}{x^2 + x + 2} \, \mathrm{d}x = \int_{1/\sqrt{7}}^{6/\sqrt{7}} \frac47 \cdot \frac{\tfrac{\sqrt{7}}{2}u + \tfrac52}{u^2 + 1} \frac{\sqrt{7}}{2} \, \mathrm{d}u = \frac{2\sqrt{7}}{7} \int_{1/\sqrt{7}}^{6/\sqrt{7}} \frac{\tfrac{\sqrt{7}}{2}u + \tfrac52}{u^2 + 1} \mathrm{d}u $$

6. Break into two pieces with an eye towards integration. $$ \frac{2\sqrt{7}}{7} \int_{1/\sqrt{7}}^{6/\sqrt{7}} \frac{\tfrac{\sqrt{7}}{2}u + \tfrac52}{u^2 + 1} \mathrm{d}u = \frac12 \int_{1/\sqrt{7}}^{6/\sqrt{7}} \frac{2u}{u^2 + 1} \mathrm{d}u + \frac{5\sqrt{7}}{7} \int_{1/\sqrt{7}}^{6/\sqrt{7}} \frac{1}{u^2 + 1} \mathrm{d}u $$

7. Find antiderivatives and evaluate. I leave this to you. Hint: you get a logarithm and an inverse tangent term.

Answer 3

Assuming you have something of the form

$$\int_a^b {Av + B \over v^2 + c} \: dv $$

you can substitute $u = v/\sqrt{c}$; the numerator will remain linear, the denominator will become $c(u^2 + 1)$, and the limits of integration will change. Then clear out the $c$ in the denominator to get your result.

Answer 4

We better split the integrand into two. $$ \begin{aligned}\int_0^1 \frac{x+3}{x^2+x+2} d x=&\frac{1}{2} \int_0^1 \frac{(2 x+1)+5}{x^2+x+2} d x \\ = & \frac{1}{2}\left[\int \frac{(2 x+1) d x}{x^2+x+2}+20 \int \frac{d x}{4 x^2+4 x+8}\right] \\ = & \frac{1}{2}\left(\left[\ln \left(x^2+x+2\right)\right]_0^1+10 \int \frac{d(2 x+1)}{(2 x+1)^2+7}\right) \\ = & \frac{1}{2} \ln 2+\frac{5}{\sqrt{7}}\left[\tan ^{-1}\left(\frac{2 x+1}{\sqrt{7}}\right)\right]_0^1 \\ = & \frac{1}{2} \ln 2+\frac{5}{\sqrt{7}}\left[\tan ^{-1}\left(\frac{3}{\sqrt{7}}\right)-\tan ^{-1}\left(\frac{1}{\sqrt{7}}\right)\right] \end{aligned} $$