# What is f(x) = int 1/((x+3)(x^2+4) dx if f(2) = 0 ?

Jun 12, 2018

$f \left(x\right) = \frac{2 \ln \left\mid x + 3 \right\mid - \ln \left({x}^{2} + 4\right) + 3 \arctan \left(\frac{x}{2}\right)}{26} - \frac{8 \ln 5 - 4 \ln 8 + 3 \pi}{104}$

#### Explanation:

$f \left(x\right) = {\int}_{2}^{x} \frac{\mathrm{dt}}{\left(t + 3\right) \left({t}^{2} + 4\right)}$

Apply partial fractions decomposition:

$\frac{1}{\left(t + 3\right) \left({t}^{2} + 4\right)} = \frac{A}{t + 3} + \frac{B t + C}{{t}^{2} + 4}$

$1 = A {t}^{2} + 4 A + B {t}^{2} + C t + 3 B t + 3 C$

$1 = \left(A + B\right) {t}^{2} + \left(C + 3 B\right) t + 3 C + 4 A$

$\left\{\begin{matrix}A + B = 0 \\ C + 3 B = 0 \\ 3 C + 4 A = 1\end{matrix}\right.$

$\left\{\begin{matrix}A = - B \\ - 3 A + C = 0 \\ 4 A + 3 C = 1\end{matrix}\right.$

$\left\{\begin{matrix}A = \frac{1}{13} \\ B = - \frac{1}{13} \\ C = \frac{3}{13}\end{matrix}\right.$

Then using the linearity of the integral:

$f \left(x\right) = \frac{1}{13} {\int}_{2}^{x} \frac{\mathrm{dt}}{t + 3} - \frac{1}{13} {\int}_{2}^{x} \frac{t \mathrm{dt}}{{t}^{2} + 4} + \frac{3}{13} {\int}_{2}^{x} \frac{\mathrm{dt}}{{t}^{2} + 4}$

Solve the integrals separately:

${\int}_{2}^{x} \frac{\mathrm{dt}}{t + 3} = {\int}_{2}^{x} \frac{d \left(t + 3\right)}{t + 3} = {\left[\ln \left\mid t + 3 \right\mid\right]}_{2}^{x} = \ln \left\mid x + 3 \right\mid - \ln 5$

${\int}_{2}^{x} \frac{t \mathrm{dt}}{{t}^{2} + 4} = \frac{1}{2} {\int}_{2}^{x} \frac{d \left({t}^{2} + 4\right)}{{t}^{2} + 4} = \frac{1}{2} {\left[\ln \left({t}^{2} + 4\right)\right]}_{2}^{x} = \ln \left({x}^{2} + 4\right) - \ln 8$

${\int}_{2}^{x} \frac{\mathrm{dt}}{{t}^{2} + 4} = \frac{1}{2} {\int}_{2}^{x} \frac{d \left(\frac{t}{2}\right)}{{\left(\frac{t}{2}\right)}^{2} + 1} = \frac{1}{2} {\left[\arctan \left(\frac{t}{2}\right)\right]}_{2}^{x} = \frac{1}{2} \left(\arctan \left(\frac{x}{2}\right) - \arctan \left(1\right)\right) = \frac{1}{2} \arctan \left(\frac{x}{2}\right) - \frac{\pi}{8}$

Putting it together:

$f \left(x\right) = \frac{2 \ln \left\mid x + 3 \right\mid - \ln \left({x}^{2} + 4\right) + 3 \arctan \left(\frac{x}{2}\right)}{26} - \frac{8 \ln 5 - 4 \ln 8 + 3 \pi}{104}$