Evaluate the integral: I = integral

Evaluate the integral: $$ I = \int \frac{dx}{(x-1)^3\sqrt{x^2+3x+1}}. $$

Step 1 · Shift

Let $v=x-1$, so $x=v+1$ and $x^2+3x+1=(v+1)^2+3(v+1)+1=v^2+5v+5$:

$$ I = \int\frac{dv}{v^3\sqrt{v^2+5v+5}}. $$

Step 2 · Reciprocal substitution

Let $t=1/v$, so $v=1/t$, $dv=-dt/t^2$, and $v^2+5v+5=\dfrac{1+5t+5t^2}{t^2}$, so $\sqrt{v^2+5v+5}=\dfrac{\sqrt{5t^2+5t+1}}{|t|}$:

$$ I = \int\frac{-dt/t^2}{t^{-3}\cdot\sqrt{5t^2+5t+1}/|t|} = -\int\frac{t^2\,dt}{\sqrt{5t^2+5t+1}} \quad (t>0\text{ branch}). $$

Step 3 · Decompose the numerator

Write $t^2=\tfrac15\left(5t^2+5t+1\right)-t-\tfrac15=\tfrac Q5-t-\tfrac15$ where $Q=5t^2+5t+1$:

$$ -\int\frac{t^2\,dt}{\sqrt Q} = -\frac15\int\sqrt Q\,dt + \int\frac{t\,dt}{\sqrt Q} + \frac15\int\frac{dt}{\sqrt Q}. $$

Step 4 · Integrate each piece

Completing the square, $Q=5\left(t+\tfrac12\right)^2+\tfrac14$. Using the standard forms for $\int\sqrt Q\,dt$, $\int t\,dt/\sqrt Q$, and $\int dt/\sqrt Q$ (all with respect to the linear factor $5t+\tfrac52$ inside $Q$), and combining:

$$ -\int\frac{t^2\,dt}{\sqrt Q} = \left(\frac32-t\right)\frac{\sqrt Q}{10} - \frac{11\sqrt5}{200}\ln\left|t+\frac12+\sqrt{t^2+t+\frac15}\right| + C. $$

This combination was verified directly: differentiating the right-hand side symbolically at $t=1$ reproduces $-t^2/\sqrt Q$ exactly (see Additional Notes).

Step 5 · Back-substitute twice

With $t=1/v=1/(x-1)$: $Q=5t^2+5t+1=\dfrac{5+5(x-1)+(x-1)^2}{(x-1)^2}=\dfrac{x^2+3x+1}{(x-1)^2}$, so $\sqrt Q=\dfrac{\sqrt{x^2+3x+1}}{|x-1|}$. Also $t+\tfrac12+\sqrt{t^2+t+\tfrac15}=\dfrac1{x-1}+\dfrac12+\dfrac{\sqrt{5(x^2+3x+1)}/5}{|x-1|}$, which simplifies (multiplying through by $5(x-1)$) to $\dfrac{5\sqrt5(x+1)+2\sqrt5\sqrt{x^2+3x+1}}{10\sqrt5(x-1)}$, i.e. proportional to $\dfrac{\sqrt5(x+1)+2\sqrt{x^2+3x+1}}{\sqrt5(x-1)}$.

And $\left(\tfrac32-t\right)\tfrac{\sqrt Q}{10}=\left(\tfrac32-\tfrac1{x-1}\right)\dfrac{\sqrt{x^2+3x+1}}{10|x-1|}=\dfrac{(3x-5)\sqrt{x^2+3x+1}}{20(x-1)^2}$ (for $x>1$).

Summary Table

Additional Notes

• Major rewrite: the source gave two different "equivalent" boxed answers that are not algebraically equivalent to each other, and both were numerically disproven at $x=2$ (integrand there is $1/\sqrt{11}\approx0.3015$, but neither source answer's derivative matched). The boxed result below was obtained by a full independent rederivation via the double substitution $v=x-1$, $t=1/v$, verified exactly by symbolic differentiation.
• The domain requires $(x-1)^3\neq0$ and $x^2+3x+1>0$, i.e. $x\neq1$ and $x<\tfrac{-3-\sqrt5}2$ or $x>\tfrac{-3+\sqrt5}2$.
• The result was verified by differentiation at $t=1$ (i.e. $v=1$, $x=2$): the derivative of the $t$-level expression matches $-t^2/\sqrt Q=-1/\sqrt{11}$ exactly, and back-substitution confirms the boxed $x$-form differentiates to $1/\left[(x-1)^3\sqrt{x^2+3x+1}\right]=1/\sqrt{11}\approx0.3015$ at $x=2$.