Paneer Definite Masala 14

Calculus Level 4

∫ 0 1 arctan ⁡ ( 1 − x + x 2 ) d x \large \int_0^1 \arctan(1-x+x^2) \, dx

The integral above has a closed form. Find the value of this closed form.

Give your answer to 3 decimal places

You may use the following approximations: log ⁡ 10 2 = 0.3010 , ln ⁡ 10 = 2.3026 \log_{10} 2 = 0.3010, \ln10 = 2.3026 .


The answer is 0.693147.

Kishore S. Shenoy by Kishore S. Shenoy , 22, India

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1 solution

Spandan Senapati
May 4, 2017

We use the fact that ∫ a r c t a n ( 1 / 1 − x + x 2 ) \int arctan(1/1-x+x^2) with the given limits equals 2 ∫ a r c t a n x 2\int arctanx .And express a r c t a n ( 1 − x + x 2 ) = π / 2 − a r c t a n ( 1 / 1 − x + x 2 ) arctan(1-x+x^2)=π/2-arctan(1/1-x+x^2) .Now one can easily integrate by using ∫ a r c t a n x = x a r c t a n x − 1 / 2 l n ( x 2 + 1 ) \int arctanx=xarctanx-1/2ln(x^2+1)

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