Answer
$$\int^{\sqrt3}_1\frac{3t^2+t+4}{t^3+t}dt=\ln\frac{9\sqrt2}{2}+\frac{\pi}{12}$$
Work Step by Step
$$A=\int^{\sqrt3}_1\frac{3t^2+t+4}{t^3+t}dt=\int^{\sqrt3}_1\frac{3t^2+t+4}{t(t^2+1)}dt$$
1) Express the integrand as a sum of partial fractions:
Since $t^2+1$ is an irreducible quadratic factor,
$$\frac{3t^2+t+4}{t(t^2+1)}=\frac{A}{t}+\frac{Bt+C}{t^2+1}$$
Clear fractions:
$$At^2+A+Bt^2+Ct=3t^2+t+4$$ $$(A+B)t^2+Ct+A=3t^2+t+4$$
Equating coefficients of corresponding powers of $t$, we get
- $A+B=3$
- $C=1$
- $A=4$
Solving for $B$, we get $B=-1$.
Therefore,
$$\frac{3t^2+t+4}{t(t^2+1)}=\frac{4}{t}+\frac{1-t}{t^2+1}$$
2) Evaluate the integral:
$$A=4\int^{\sqrt3}_1\frac{dt}{t}+\int^{\sqrt3}_1\frac{dt}{t^2+1}-\int^{\sqrt3}_1\frac{t}{t^2+1}dt$$ $$A=4\ln|t|\Big]^{\sqrt3}_1+\tan^{-1}t\Big]^{\sqrt3}_1-\frac{1}{2}\int^{\sqrt3}_1\frac{1}{t^2+1}d(t^2+1)$$ $$A=4\ln\sqrt3+\tan^{-1}\sqrt3-\tan^{-1}1-\frac{1}{2}\ln|t^2+1|\Big]^{\sqrt3}_1$$ $$A=4\ln\sqrt3+\frac{\pi}{3}-\frac{\pi}{4}-\frac{1}{2}(\ln4-\ln2)$$ $$A=\ln9-\frac{1}{2}\ln2+\frac{\pi}{12}$$ $$A=\ln9-\ln\sqrt2+\frac{\pi}{12}$$ $$A=\ln\frac{9\sqrt2}{2}+\frac{\pi}{12}$$
