Editing Openai/6936f996-e124-8004-a264-0b3da0e38976


{{DISPLAYTITLE:Analytical integration solution}}

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== gpt-5-1-thinking ==
=== User: Can it be taken analytically? ===
Can it be taken analytically?

=== Assistant: Yes, it can be integrated in closed form. ===
Yes, it can be integrated in closed form.

∫3x3−x2+2x−4x2−3x+2 dx=x2−3x+2(x2+134x+1018)+13516ln⁡ ⁣∣2x−3+2x2−3x+2∣+C.\int \frac{3x^{3}-x^{2}+2x-4}{\sqrt{x^{2}-3x+2}}\,dx
= \sqrt{x^{2}-3x+2}\left(x^{2}+\frac{13}{4}x+\frac{101}{8}\right)
+\frac{135}{16}\ln\!\left|2x-3+2\sqrt{x^{2}-3x+2}\right|+C.∫x2−3x+23x3−x2+2x−4dx=x2−3x+2(x2+413x+8101)+16135ln2x−3+2x2−3x+2+C.
Sketch of how to get it:
# Denote Q(x)=x2−3x+2Q(x)=x^{2}-3x+2Q(x)=x2−3x+2. Look for an antiderivative of the form Q(x)(Ax2+Bx+C)+KQ(x)\sqrt{Q(x)}(Ax^{2}+Bx+C)+\dfrac{K}{\sqrt{Q(x)}}Q(x)(Ax2+Bx+C)+Q(x)K.
# Differentiate this expression, collect terms over Q(x)\sqrt{Q(x)}Q(x), and match the resulting polynomial in the numerator with 3x3−x2+2x−43x^{3}-x^{2}+2x-43x3−x2+2x−4. Solving for A,B,C,KA,B,C,KA,B,C,K gives A=1,  B=134,  C=1018,  K=13516A=1,\;B=\tfrac{13}{4},\;C=\tfrac{101}{8},\;K=\tfrac{135}{16}A=1,B=413,C=8101,K=16135.
# The remaining integral ∫dxx2−3x+2\displaystyle \int \frac{dx}{\sqrt{x^{2}-3x+2}}∫x2−3x+2dx is handled by completing the square x2−3x+2=(x−32)2−(12)2x^{2}-3x+2=(x-\tfrac32)^2-(\tfrac12)^2x2−3x+2=(x−23)2−(21)2 and using ∫dxx2−a2=ln⁡∣x+x2−a2∣+C.\displaystyle \int \frac{dx}{\sqrt{x^{2}-a^{2}}}    =\ln\left|x+\sqrt{x^{2}-a^{2}}\right|+C.∫x2−a2dx=lnx+x2−a2+C.

Combining gives the result above.

=== User: So, what is the analytical answer, and then calculate the answer from the analytical answer to get number. ===
So, what is the analytical answer, and then calculate the answer from the analytical answer to get number.

=== Assistant: Let’s write the integral clearly: ===
Let’s write the integral clearly:

∫3x3−x2+2x−4x2−3x+2 dx\int \frac{3x^{3}-x^{2}+2x-4}{\sqrt{x^{2}-3x+2}}\,dx∫x2−3x+23x3−x2+2x−4dx
===== 1. Analytical antiderivative =====

One antiderivative is

F(x)=x2−3x+2(x2+134x+1018)+13516ln⁡ ⁣∣2x−3+2x2−3x+2∣+C.F(x)=\sqrt{x^{2}-3x+2}\left(x^{2}+\frac{13}{4}x+\frac{101}{8}\right)
+\frac{135}{16}\ln\!\left|2x-3+2\sqrt{x^{2}-3x+2}\right|+C.F(x)=x2−3x+2(x2+413x+8101)+16135ln2x−3+2x2−3x+2+C.
(Differentiating F(x)F(x)F(x) gives exactly the integrand.)

===== 2. Definite integral (taking limits =====
0
0 to 
1
1)

Your picture shows a lower limit 000; the natural upper limit, given the domain of (x−1)(x−2)\sqrt{(x-1)(x-2)}(x−1)(x−2), is 111.
Assuming the integral is

∫013x3−x2+2x−4x2−3x+2 dx,\int_{0}^{1} \frac{3x^{3}-x^{2}+2x-4}{\sqrt{x^{2}-3x+2}}\,dx,∫01x2−3x+23x3−x2+2x−4dx,
we plug into F(x)F(x)F(x):

∫013x3−x2+2x−4x2−3x+2 dx=F(1)−F(0)=−10182−13516ln⁡ ⁣(3−22).\int_{0}^{1} \frac{3x^{3}-x^{2}+2x-4}{\sqrt{x^{2}-3x+2}}\,dx
=F(1)-F(0)
=-\frac{101}{8}\sqrt{2}-\frac{135}{16}\ln\!\bigl(3-2\sqrt{2}\bigr).∫01x2−3x+23x3−x2+2x−4dx=F(1)−F(0)=−81012−16135ln(3−22).
Numerically (to 10 digits):

∫01⋯dx≈−2.981266944.\int_{0}^{1} \cdots dx \approx -2.981266944.∫01⋯dx≈−2.981266944.
Confidence this is correct: ~99% (symbolically checked by differentiating the antiderivative and by direct numeric integration).

=== User: I did not ask numerically integrate. ===
I did not ask numerically integrate. Calculate from your analytical answer to definite integral.

=== Assistant: Let me assume the definite integral is ===
Let me assume the definite integral is

I=∫013x3−x2+2x−4x2−3x+2 dx,I=\int_{0}^{1} \frac{3x^{3}-x^{2}+2x-4}{\sqrt{x^{2}-3x+2}}\,dx ,I=∫01x2−3x+23x3−x2+2x−4dx,
since x2−3x+2=(x−1)(x−2)\sqrt{x^{2}-3x+2}=\sqrt{(x-1)(x-2)}x2−3x+2=(x−1)(x−2) is real on [0,1][0,1][0,1].

===== 1. Analytical evaluation =====

An antiderivative is

F(x)=x2−3x+2(x2+134x+1018)+13516ln⁡ ⁣(2x−3+2x2−3x+2).F(x)=\sqrt{x^{2}-3x+2}\left(x^{2}+\frac{13}{4}x+\frac{101}{8}\right)
+\frac{135}{16}\ln\!\bigl(2x-3+2\sqrt{x^{2}-3x+2}\bigr).F(x)=x2−3x+2(x2+413x+8101)+16135ln(2x−3+2x2−3x+2).
So

I=F(1)−F(0).I = F(1)-F(0).I=F(1)−F(0).
* At x=1x=1x=1: 1−3+2=0\sqrt{1-3+2}=01−3+2=0, and the log term is ln⁡(0)\ln(0)ln(0), which → −∞-\infty−∞, but multiplied by 0 from the prefactor in the derivation the limit is finite; easier is to use the symbolic simplification already done, which gives the exact result below.

Carrying out the algebra (which SymPy also confirms) gives

I=−10182−13516ln⁡(3−22).I=-\frac{101}{8}\sqrt{2}-\frac{135}{16}\ln(3-2\sqrt{2}).I=−81012−16135ln(3−22).
Using (3−22)(3+22)=1⇒ln⁡(3−22)=−ln⁡(3+22)(3-2\sqrt{2})(3+2\sqrt{2})=1\Rightarrow
\ln(3-2\sqrt{2})=-\ln(3+2\sqrt{2})(3−22)(3+22)=1⇒ln(3−22)=−ln(3+22), a slightly nicer exact form is

 I=−10182+13516ln⁡(3+22) .\boxed{\,I=-\frac{101}{8}\sqrt{2}+\frac{135}{16}\ln(3+2\sqrt{2})\,}.I=−81012+16135ln(3+22).
That is the analytical answer.

===== 2. Numerical value from this formula (no numerical integration) =====

Use

2≈1.4142135623730950,ln⁡(3+22)≈1.7627471740390861.\sqrt{2}\approx 1.4142135623730950,\qquad
\ln(3+2\sqrt{2})\approx 1.7627471740390861.2≈1.4142135623730950,ln(3+22)≈1.7627471740390861.
Compute the two terms:

−10182≈−12.625⋅1.4142135623730950≈−17.854446224960325,-\frac{101}{8}\sqrt{2}
\approx -12.625\cdot 1.4142135623730950
\approx -17.854446224960325,−81012≈−12.625⋅1.4142135623730950≈−17.854446224960325,
13516ln⁡(3+22)=8.4375⋅1.7627471740390861≈14.873179280954789.\frac{135}{16}\ln(3+2\sqrt{2})
=8.4375\cdot 1.7627471740390861
\approx 14.873179280954789.16135ln(3+22)=8.4375⋅1.7627471740390861≈14.873179280954789.
Add them:

I≈−17.854446224960325+14.873179280954789≈−2.981266944005536.I \approx -17.854446224960325
+14.873179280954789
\approx -2.981266944005536.I≈−17.854446224960325+14.873179280954789≈−2.981266944005536.
So

I≈−2.9812669440\boxed{I\approx -2.9812669440}I≈−2.9812669440
(from the closed-form expression, accuracy ≳10 significant digits).

== References ==
<references />


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