Integration 1

Calculus Level 3

∫ 1 2 x ( x 2 + 1 ) ( x 2 + 2 ) d x \large \int_1^2 \dfrac x{(x^2+1)(x^2+2)} \, dx

If the integral above can be expressed as ln ⁡ a b \ln \sqrt{\dfrac ab } , where a a and b b are coprime positive integers, find a + b a+b .


The answer is 9.

Akshay Sharma by Akshay Sharma , 21, India

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1 solution

Chew-Seong Cheong
Mar 12, 2016

I = ∫ 1 2 x ( x 2 + 1 ) ( x 2 + 2 ) d x Let u = x 2 ⇒ d u = 2 x d x = 1 2 ∫ 1 4 1 ( u + 1 ) ( u + 2 ) d u = 1 2 ∫ 1 4 ( 1 u + 1 − 1 u + 2 ) d u = 1 2 [ ln ⁡ ( u + 1 ) − ln ⁡ ( u + 2 ) ] 1 4 = 1 2 [ ln ⁡ ( u + 1 u + 2 ) ] 1 4 = 1 2 [ ln ⁡ ( 5 6 ) − ln ⁡ ( 2 3 ) ] = 1 2 ln ⁡ ( 5 4 ) = ln ⁡ 5 4 \begin{aligned} I & = \int_1^2 \frac{x}{(x^2+1)(x^2+2)}dx \quad \quad \small \color{#3D99F6}{\text{Let }u = x^2 \quad \Rightarrow du = 2 x\space dx} \\ & = \frac{1}{2} \int_1^4 \frac{1}{(u+1)(u+2)}du \\ & = \frac{1}{2} \int_1^4 \left(\frac{1}{u+1} - \frac{1}{u+2} \right) du \\ & = \frac{1}{2} \left[ \ln(u+1) - \ln(u+2) \right]_1^4 \\ & = \frac{1}{2} \left[ \ln \left(\frac{u+1}{u+2}\right) \right]_1^4 \\ & = \frac{1}{2} \left[ \ln \left(\frac{5}{6}\right) - \ln \left(\frac{2}{3} \right) \right] \\ & = \frac{1}{2} \ln \left(\frac{5}{4}\right) = \ln \sqrt{\frac{5}{4}} \end{aligned}

⇒ a + b = 5 + 4 = 9 \Rightarrow a + b = 5 + 4 = \boxed{9}

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Your solutions are always very instructive. Thank you!

Paulo Filho - 5 years, 3 months ago

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Thanks. I always try to do my best.

Chew-Seong Cheong - 5 years, 3 months ago

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