This is quite easy

Calculus Level pending

The integral ∫ 0 1 2 x + 3 x 2 + 2 x + 3 d x \displaystyle\int_{0}^{1}\frac{2x+3}{x^{2}+2x+3}dx can be written as l n ( a ) + b ( t a n − 1 ( c ) + t a n − 1 ( d ) ) ln(a)+b(tan^{-1}(c)+tan^{-1}(d)) . Find ( a b c d ) 2 − 1 (abcd)^{2}-1


The answer is 1.

Hjalmar Orellana Soto by Hjalmar Orellana Soto , 24, Chile

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1 solution

∫ 0 1 2 x + 3 x 2 + 2 x + 3 d x \displaystyle\int_{0}^{1}\frac{2x+3}{x^{2}+2x+3}dx = ∫ 0 1 2 x + 2 x 2 + 2 x + 3 d x + ∫ 0 1 d x ( x + 1 ) 2 + 2 =\displaystyle\int_{0}^{1}\frac{2x+2}{x^{2}+2x+3}dx+\displaystyle\int_{0}^{1}\frac{dx}{(x+1)^{2}+2} l n ( 6 ) − l n ( 3 ) + 1 2 ∫ 0 1 d x ( x + 1 2 ) 2 + 1 ln(6)-ln(3)+\frac{1}{2}\displaystyle\int_{0}^{1}\frac{dx}{(\frac{x+1}{\sqrt{2}})^{2}+1} l n ( 2 ) + 1 2 ( t a n − 1 ( 2 ) − t a n − 1 ( 1 2 ) ) ln(2)+\frac{1}{2}(tan^{-1}(\sqrt{2})-tan^{-1}(\frac{1}{\sqrt{2}}))

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