Three integrals: by parts, partial fractions, completing the square

Text

Evaluate, using respectively integration by parts, partial fractions and completing the square: a) $\int_0^{\pi/2}x\cos x\,dx$ b) $\int_2^3\dfrac{3x-1}{(x-1)(x+1)}\,dx$ c) $\int_0^1\dfrac{dx}{x^2+4x+5}$

Solution

a) By parts ($u=x$, $dv=\cos x\,dx$): $\int x\cos x\,dx=x\sin x+\cos x$. $\big[x\sin x+\cos x\big]_0^{\pi/2}=\left(\frac{\pi}{2}\cdot 1+0\right)-(0+1)=\frac{\pi}{2}-1\approx0.5708.$

b) Partial fractions. $\dfrac{3x-1}{(x-1)(x+1)}=\dfrac{A}{x-1}+\dfrac{B}{x+1}$ with $3x-1=A(x+1)+B(x-1)$: $x=1\Rightarrow A=1$; $x=-1\Rightarrow B=2$. $\int_2^3\!\left(\frac{1}{x-1}+\frac{2}{x+1}\right)dx=\big[\ln|x-1|+2\ln|x+1|\big]_2^3=\ln 2+2\ln\frac43\approx1.2685.$

c) Completing the square. $x^2+4x+5=(x+2)^2+1$: $\int_0^1\frac{dx}{(x+2)^2+1}=\big[\arctan(x+2)\big]_0^1=\arctan 3-\arctan 2\approx0.1419.$

$\boxed{a=\frac{\pi}{2}-1,\quad b=\ln 2+2\ln\tfrac43,\quad c=\arctan 3-\arctan 2}$