Question

Evaluate the definite integrals: $\int^2_0 {\frac{6x+3}{x^2 +4}\, dx$

Answer 1

Topic:

Integration

Solution:

We have to solve the given integral.

${\displaystyle \int_0^2 \dfrac{6x + 3}{x^2 + 4}\, dx}$

We can break the fraction into two fractions.

${\displaystyle\longrightarrow \int_0^2 \dfrac{6x }{x^2 + 4} + \dfrac{3}{x^2 + 4}\, dx}$

${\displaystyle\longrightarrow \int_0^2 \dfrac{6x }{x^2 + 4}\, dx + \int_0^2\dfrac{3}{x^2 + 4}\, dx}$

${\displaystyle\longrightarrow 3\int_0^2\dfrac{2x }{x^2 + 4}\, dx +3 \int_0^2\dfrac{dx}{x^2 + 2^2}}$

In both integrals, we can use the following identities:

$\boxed{\int \dfrac{f'(x) \, dx}{f(x)} = \log|f(x)| + C}$

$\boxed{\int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a} \tan^{-1}\left(\dfrac{x}{a}\right) + C}$

Using the above formulas in integral, we get:

${\longrightarrow 3\bigg[\log|x^2 + 4|\bigg]^2_0 +3 \bigg[\dfrac{1}{2}\tan^{-1}\left(\dfrac{x}{2}\right)\bigg]^2_0}$

${\longrightarrow 3\bigg[\log|2^2 + 4| - \log|0^2 + 4|\bigg] +3 \bigg[\dfrac{1}{2}\tan^{-1}\left(\dfrac{2}{2}\right) - \dfrac{1}{2}\tan^{-1}\left(\dfrac{0}{2}\right)\bigg]}$

${\longrightarrow 3\bigg[\log(8) - \log(4)\bigg] +3 \bigg[\dfrac{1}{2}\tan^{-1}\left(1\right) - \dfrac{1}{2}\tan^{-1}\left(0\right)\bigg]}$

${\longrightarrow 3\bigg[\log\left(\dfrac84\right)\bigg] +3 \bigg[\dfrac{1}{2}\cdot\dfrac{\pi}{4} - 0\bigg]\qquad\qquad\left\{\because \log\left(\dfrac ab\right) \iff \log(a) - \log(b)\right\}}$

${\longrightarrow 3\log(2) + \dfrac{3\pi}{8}}$

${\displaystyle\longrightarrow \log(2^3) + \dfrac{3\pi}{8}\qquad\qquad\left\{\because b\log(a) \iff \log(a^b)\right\}}$

${\longrightarrow \log(8) + \dfrac{3\pi}{8}}$

Hence this is the required answer.