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Evaluate the definite integrals in Exercises 1 to 20.

    Q10.    \int_0^1\frac{dx}{1 + x^2}

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Given integral: \int_0^1\frac{dx}{1 + x^2}

Consider the integral \int\frac{dx}{1 + x^2}

\int\frac{dx}{1 + x^2} = \tan^{-1}x

So, we have the function of xf(x) =\tan^{-1}x

Now, by Second fundamental theorem of calculus, we have

I = f(1) -f(0)

= \tan^{-1}(1) -\tan^{-1}(0)

= \frac{\pi}{4} - 0

= \frac{\pi}{4}

Posted by

Divya Prakash Singh

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