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

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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+2​3x3−x2+2x−4​dx=x2−3x+2​(x2+413​x+8101​)+16135​ln​2x−3+2x2−3x+2​​+C. Sketch of how to get it:

1. 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​.
2. 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​.
3. The remaining integral ∫dxx2−3x+2\displaystyle \int \frac{dx}{\sqrt{x^{2}-3x+2}}∫x2−3x+2​dx​ 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−a2​dx​=ln​x+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+2​3x3−x2+2x−4​dx

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+413​x+8101​)+16135​ln​2x−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,∫01​x2−3x+2​3x3−x2+2x−4​dx, 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).∫01​x2−3x+2​3x3−x2+2x−4​dx=F(1)−F(0)=−8101​2​−16135​ln(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=∫01​x2−3x+2​3x3−x2+2x−4​dx, 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+413​x+8101​)+16135​ln(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=−8101​2​−16135​ln(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=−8101​2​+16135​ln(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,−8101​2​≈−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 14.873179280954789.16135​ln(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).