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  • Evaluate the definite integrals in Exercises 1 to 20. Integral 0 to 2 6x + 3 /x^2 + 4 dx

Evaluate the definite integrals in Exercises 1 to 20.

    Q19.    \int_0^2\frac{6x+3}{x^2+ 4}

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Given integral: \int_0^2\frac{6x+3}{x^2+ 4}

Consider the integral \int \frac{6x+3}{x^2+ 4}

can be rewritten as: \int \frac{6x+3}{x^2+ 4} = 3\int \frac{2x+1}{x^2+4}dx

= 3\int \frac{2x}{x^2+4}dx +3\int \frac{1}{x^2+4}dx

= 3\log (x^2+4) +\frac{3}{2}\tan^{-1}\frac{x}{2}

 So, we have the function of x, f(x) =3\log (x^2+4) +\frac{3}{2}\tan^{-1}\frac{x}{2}

Now, by Second fundamental theorem of calculus, we have

I = f(2) - f(0)

= \left \{ 3\log(2^2+4)+\frac{3}{2}\tan^{-1}\left ( \frac{2}{2} \right ) \right \}- \left \{ 3\log(0+4)+\frac{3}{2}\tan^{-1}\left ( \frac{0}{2} \right ) \right \}=3\log 8 +\frac{3}{2}\tan^{-1}1 -3\log 4 -\frac{3}{2}\tan^{-1} 0

=3\log 8 +\frac{3}{2}\times\frac{\pi}{4} -3\log 4 -0

=3\log \frac{8}{4} +\frac{3\pi}{8}

or we have =3\log 2 +\frac{3\pi}{8}

Posted by

Divya Prakash Singh

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