Answer to Question #185295 in Math for Gatare james

Question

a) ∫𝝅 𝒙𝒔𝒊𝒏𝟐𝒙 𝒅𝒙

𝟎

b)∫𝒙𝟐 𝒅𝒙 (𝒙+𝟏)

c)∫ 𝒙+𝟏 𝒅𝒙 𝒙𝟐 −𝟑𝒙+𝟐

Answer 1

a)

$$\displaystyle\int_{0}^\pi x\sin(2x)dx$$

$$\int x\sin(2x)dx$$

$$\int udv=uv-\int v du$$

$$u=x, du=dx$$

$$dv=\sin(2x)dx, v=\int\sin(2x)dx=-\dfrac{1}{2}\cos(2x)$$

$$\int x\sin(2x)dx=-\dfrac{1}{2}x\cos(2x)+\dfrac{1}{2}\int\cos(2x)dx$$

$$=-\dfrac{1}{2}x\cos(2x)+\dfrac{1}{4}\sin(2x)+C$$

$$\displaystyle\int_{0}^\pi x\sin(2x)dx=\big[-\dfrac{1}{2}x\cos(2x)+\dfrac{1}{4}\sin(2x)\big]\begin{matrix} \pi \\ 0 \end{matrix}$$

$$=-\dfrac{1}{2}\pi+0-(-0+0)=-\dfrac{\pi}{2}$$

b)

$$\int\dfrac{x^2}{x+1}dx=\int\dfrac{x^2+x-x-1+1}{x+1}dx$$

$$=\int xdx-\int dx+\int \dfrac{1}{x+1}dx$$

$$=\dfrac{x^2}{2}-x+\ln|x+1|+C$$

c)

$$\int\dfrac{x+1}{x^2-3x+2}dx=\int\dfrac{x+1}{(x-1)(x-2)}dx$$

$$\dfrac{x+1}{(x-1)(x-2)}=\dfrac{A}{x-1}+\dfrac{B}{x-2}$$

$$=\dfrac{A(x-2)+B(x-1)}{(x+1)(x-2)}$$

$x=1: A=-2$

$x=2: B=3$

$$\int\dfrac{x+1}{x^2-3x+2}dx=-2\int \dfrac{1}{x-1}dx+3\int \dfrac{1}{x-2}dx$$

$$=-2\ln|x-1|+3\ln|x-2|+C$$