Exercise 10.4

Exercise 10.4

1. Use the partial fractions method to evaluate the following integrals.

(a) $\int \frac{1}{2x^2+5x+3}dx$

Solution:

$$ \frac{1}{2x^2+5x+3} = \frac{1}{(2x+3)(x+1)} = \frac{A}{2x+3} + \frac{B}{x+1} $$

$$ 1 = A(x+1) + B(2x+3) = (A+2B)x + (A+3B) $$

By equating the coefficients of corresponding powers of x,

$$ A+2B=0 \tag{1} $$ $$ A+3B=1 \tag{2} $$

If eq(2)-eq(1), $B=1$.

If $B=1$ in eq(1), $A=-2$.

$$ \begin{aligned} \text{Therefore, } \int \frac{1}{(2x+3)(x+1)}dx &= \int (\frac{-2}{2x+3} + \frac{1}{x+1})dx \\ &= \ln|x+1| - \ln|2x+3| + C \\ &= \ln|\frac{x+1}{2x+3}| + C \end{aligned} $$

(b) $\int \frac{2x-1}{(x-3)^2}dx$

Solution:

$$ \frac{2x-1}{(x-3)^2} = \frac{A}{x-3} + \frac{B}{(x-3)^2} $$

$$ 2x-1 = A(x-3) + B = Ax - (3A-B) $$

By equating the coefficients of corresponding powers of x, $A=2$

$$ 3A-B=1 \tag{1} $$

If $A=2$ in eq (1), $B=5$.

$$ \begin{aligned} \text{Therefore, } \int \frac{2x-1}{(x-3)^2}dx &= \int (\frac{2}{x-3} + \frac{5}{(x-3)^2})dx \\ &= 2 \int \frac{1}{x-3}dx + 5 \int(x-3)^{-2}dx \\ &= 2\ln|x-3| - \frac{5}{x-3} + C \end{aligned} $$

(c) $\int \frac{x+1}{(2x+5)(x+4)}dx$

Solution:

$$ \frac{x+1}{(2x+5)(x+4)} = \frac{A}{2x+5} + \frac{B}{x+4} $$

$$ x+1 = A(x+4) + B(2x+5) = (A+2B)x + (4A+5B) $$

By equating coefficients:

$$ A+2B=1 \tag{1} $$ $$ 4A+5B=1 \tag{2} $$

If eq(2)-eq(1)x4, $B=-1$. If $B=-1$ in eq(1), $A=1$.

$$ \begin{aligned} \text{Therefore, } \int \frac{x+1}{(2x+5)(x+4)}dx &= \int (\frac{1}{2x+5} - \frac{1}{x+4})dx \\ &= \frac{1}{2}\ln|2x+5| - \ln|x+4| + C \\ &= \ln \frac{\sqrt{|2x+5|}}{|x+4|} + C \end{aligned} $$

(d) $\int \frac{2x^2-1}{x^2-1}dx$

Solution:

First, we write $$ \int \frac{2x^2-1}{x^2-1}dx = \int (2 + \frac{1}{x^2-1})dx $$

First, we write $$ \frac{1}{x^2-1} = \frac{1}{(x+1)(x-1)} = \frac{A}{x+1} + \frac{B}{x-1} $$

and so $$ 1 = A(x-1) + B(x+1) $$

When $x=1$, $1=B(2)$, so $B=1/2$.

When $x=-1$, $1=A(-2)$, so $A=-1/2$.

$$ \begin{aligned} \text{Therefore, } \int \frac{2x^2-1}{x^2-1}dx &= \int 2 dx - \int \frac{\frac{1}{2}}{x+1}dx + \int \frac{\frac{1}{2}}{x-1}dx \\ &= 2x - \frac{1}{2}\ln|x+1| + \frac{1}{2}\ln|x-1| + C \\ &= 2x + \frac{1}{2}\ln|\frac{x-1}{x+1}| + C \end{aligned} $$

2. Find the function $f(x)$ that satisfying the equation $f'(x) = \sin 4x \cos 2x$ with $f(\frac{\pi}{2}) = 0$.

Solution:

By integrating, $$ f(x) = \int \sin 4x \cos 2x dx = \frac{1}{2} \int (\sin 6x + \sin 2x) dx $$

$$ = \frac{1}{2} (-\frac{1}{6}\cos 6x - \frac{1}{2}\cos 2x) + C $$

Since $f(\frac{\pi}{2}) = 0$:

$$ \frac{1}{2} (-\frac{1}{6}\cos 6(\frac{\pi}{2}) - \frac{1}{2}\cos 2(\frac{\pi}{2})) + C = 0 $$

$$ -\frac{1}{12}\cos 3\pi - \frac{1}{4}\cos\pi + C = 0 $$

$$ C = \frac{1}{12}\cos 3\pi + \frac{1}{4}\cos\pi = \frac{1}{12}(-1) + \frac{1}{4}(-1) = -\frac{1}{3} $$

Therefore, $$ f(x) = -\frac{1}{12}\cos 6x - \frac{1}{4}\cos 2x - \frac{1}{3} $$

3. Find the function $g(x)$ that satisfying the equation $g'(x) = x^2 e^{x^3}$ with $g(0) = -\frac{2}{3}$.

Solution:

By integrating, $$ g(x) = \int x^2 e^{x^3} dx $$

Let $u = x^3$, so $\frac{1}{3}du = x^2 dx$.

$$ g(x) = \int \frac{1}{3}e^u du = \frac{1}{3} e^u + C = \frac{1}{3} e^{x^3} + C $$

Since $g(0) = -\frac{2}{3}$:

$$ \frac{1}{3} e^{0} + C = -\frac{2}{3} \text{, so } \frac{1}{3} + C = -\frac{2}{3} \text{, so } C = -1 $$

Therefore, $$ g(x) = \frac{1}{3}e^{x^3} - 1 $$

4. Find the function $h(x)$, that satisfying the equation $h'(x) = \frac{x}{x^2-1}$ with $h(2) = \frac{1}{2}$.

Solution:

By integrating, $$ h(x) = \int \frac{x}{x^2-1} dx $$

Let $u = x^2-1$, so $\frac{1}{2}du = x dx$.

$$ h(x) = \int \frac{1}{2u} du = \frac{1}{2}\ln|u| + C = \frac{1}{2}\ln|x^2-1| + C $$

Since $h(2) = \frac{1}{2}$:

$$ \frac{1}{2}\ln|2^2-1| + C = \frac{1}{2} \text{, so } \frac{1}{2}\ln|3| + C = \frac{1}{2} $$

$$ \ln|3| + 2C = 1 \text{, so } C = \frac{1 - \ln 3}{2} $$

Therefore, $$ h(x) = \frac{1}{2}\ln|x^2-1| + \frac{1 - \ln 3}{2} $$

5. Find the function $f(x)$ satisfying the equation $f''(x) = 2x - 1$ with $f(0) = -1$ and $f'(1) = 2$.

Solution:

By integrating, $$ f'(x) = \int (2x-1)dx = x^2 - x + C $$

Since $f'(1) = 2$: $1^2 - 1 + C = 2$, so $C = 2$.

Therefore $f'(x) = x^2 - x + 2$.

By integrating, we get $$ f(x) = \int (x^2 - x + 2)dx = \frac{x^3}{3} - \frac{x^2}{2} + 2x + C $$

Since $f(0) = -1$: $\frac{0}{3} - \frac{0}{2} + 2(0) + C = -1$, so $C = -1$.

Therefore, $$ f(x) = \frac{x^3}{3} - \frac{x^2}{2} + 2x - 1 $$

6. Find the function $g(x)$ that satisfying the equation $g''(x) = x \sin x$ with $g(\frac{\pi}{2}) = 0$ and $g'(0) = 0$.

Solution:

By integrating, $$ g'(x) = \int x \sin x dx $$

Let $u=x$, $dv = \sin x dx$, then $du=dx$, $v=-\cos x$.

$$ g'(x) = -x \cos x - \int (-\cos x) dx = -x \cos x + \sin x + C $$

Since $g'(0) = 0$: $-(0) \cos 0 + \sin 0 + C = 0$, so $C = 0$.

Therefore, $g'(x) = -x \cos x + \sin x$.

By integrating, $$ g(x) = \int (-x \cos x + \sin x)dx = - \int x \cos x dx + \int \sin x dx $$

Let $u=x$, $dv = \cos x dx$, then $du=dx$, $v=\sin x$.

$$ \therefore g(x) = -(x \sin x - \int \sin x dx) + \int \sin x dx = -x \sin x + 2 \int \sin x dx $$

$$ = -x \sin x - 2\cos x + C $$

Since $g(\frac{\pi}{2}) = 0$:

$$ -\frac{\pi}{2}\sin(\frac{\pi}{2}) - 2\cos(\frac{\pi}{2}) + C = 0 \text{, so } -\frac{\pi}{2}(1) - 2(0) + C = 0 \text{, so } C = \frac{\pi}{2} $$

Therefore, $$ g(x) = -x \sin x - 2 \cos x + \frac{\pi}{2} $$