It's Integration Time

Calculus Level 1

Evaluate ∫ x 3 + 1 x + 2 d x \large{\int \dfrac{x^3 + 1}{x + 2} dx}

The answer can be expressed as a x 3 − b x 2 + c x − d ln ⁡ ( x + e ) + C ax^3-bx^2+cx-d \ln (x+e)+C , where: a + b + c + d + e = x y a+b+c+d+e=\dfrac{x}{y} .

Give your answer as x + y x+y .


The answer is 46.

Marvin Kalngan by Marvin Kalngan , 38, Philippines

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1 solution

Marvin Kalngan
May 6, 2020

Divide x 3 + 1 x^3+1 by x + 2 x+2 by long division. The quotient is x 2 − 2 x + 4 x^2-2x+4 , and the remainder is − 7 -7 . So,

∫ x 3 + 1 x + 2 d x \int \dfrac{x^3 + 1}{x + 2} dx

= ∫ ( x 2 − 2 x + 4 − 7 x + 2 ) d x =\int \left(x^2-2x+4-\dfrac{7}{x+2}\right)dx

= x 3 3 − x 2 + 4 x − 7 ln ⁡ ( x + 2 ) + C =\dfrac{x^3}{3}-x^2+4x-7\ln (x+2)+C

The required answer is:

a + b + c + d + e = 1 3 + 1 + 4 + 7 + 2 = 43 3 = x y a+b+c+d+e=\dfrac{1}{3}+1+4+7+2=\dfrac{43}{3}=\dfrac{x}{y}

x + y = 43 + 3 = 46 x+y=43+3=\boxed{46}

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