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f(x)=(2x+1)(x^2-3x+4)^(1/2)−6−5−4−3−2−10123456−3-2.5−2-1.5−1-0.500.511.522.53xy

Exercise

∫(2x+1)x2−3x+4dx\int\left(2x+1\right)\sqrt{x^2-3x+4}dx

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Final answer to the exercise

23((x−32)2+74)3+7(27(x−32)(x−32)2+74+12ln⁡∣2(x−32)2+74+2x−37∣)+C0\frac{2}{3}\sqrt{\left(\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\right)^{3}}+7\left(\frac{2}{7}\left(x-\frac{3}{2}\right)\sqrt{\left(x-\frac{3}{2}\right)^2+\frac{7}{4}}+\frac{1}{2}\ln\left|\frac{2\sqrt{\left(x-\frac{3}{2}\right)^2+\frac{7}{4}}+2x-3}{\sqrt{7}}\right|\right)+C_0

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∫(2x+1)√x2−3x+4dx
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