Answer :
Certainly! Let's solve each integral step-by-step.
### (a) [tex]\(\int \frac{d x}{\sqrt{x^2-16}}\)[/tex]
1. Identify the integral form:
The integral [tex]\(\int \frac{d x}{\sqrt{x^2-16}}\)[/tex] resembles a standard integral that can be solved using a trigonometric substitution.
2. Trigonometric substitution:
Consider [tex]\( x = 4 \sec(\theta) \)[/tex], then [tex]\( dx = 4 \sec(\theta) \tan(\theta) d\theta \)[/tex].
3. Substitute and simplify:
Substitute [tex]\( x = 4 \sec(\theta) \)[/tex]:
[tex]\[ \sqrt{x^2 - 16} = \sqrt{16 \sec^2(\theta) - 16} = \sqrt{16 (\sec^2(\theta) - 1)} = 4 \tan(\theta) \][/tex]
So, the integral becomes:
[tex]\[ \int \frac{4 \sec(\theta) \tan(\theta) d\theta}{4 \tan(\theta)} = \int \sec(\theta) d\theta \][/tex]
4. Integrate:
The integral of [tex]\(\sec(\theta)\)[/tex] is [tex]\(\ln|\sec(\theta) + \tan(\theta)| + C\)[/tex].
[tex]\[ \int \sec(\theta) d\theta = \ln|\sec(\theta) + \tan(\theta)| + C \][/tex]
5. Back-substitute [tex]\( \theta \)[/tex]:
Since [tex]\( \sec(\theta) = \frac{x}{4} \)[/tex] and [tex]\( \tan(\theta) = \sqrt{\sec^2(\theta) - 1} = \sqrt{\left(\frac{x}{4}\right)^2 - 1} = \frac{\sqrt{x^2 - 16}}{4} \)[/tex],
[tex]\[ \sec(\theta) + \tan(\theta) = \frac{x}{4} + \frac{\sqrt{x^2 - 16}}{4} = \frac{x + \sqrt{x^2 - 16}}{4} \][/tex]
So, the integral becomes:
[tex]\[ \int \frac{d x}{\sqrt{x^2-16}} = \ln\left| \frac{x + \sqrt{x^2 - 16}}{4} \right| + C \][/tex]
For simplicity, we can remove the constant division inside the logarithm term:
[tex]\[ \int \frac{d x}{\sqrt{x^2-16}} = \ln|x + \sqrt{x^2 - 16}| + C \][/tex]
Therefore, the result for part (a) is:
[tex]\[ \boxed{\ln|x + \sqrt{x^2 - 16}| + C} \][/tex]
### (b) [tex]\(\int \frac{3 x - 5}{x^2 - 3 x + 2} d x\)[/tex]
1. Factor the denominator:
The denominator [tex]\(x^2 - 3x + 2\)[/tex] can be factored as [tex]\((x-1)(x-2)\)[/tex].
[tex]\[ \int \frac{3 x - 5}{(x-1)(x-2)} d x \][/tex]
2. Partial fraction decomposition:
We decompose [tex]\(\frac{3 x - 5}{(x-1)(x-2)}\)[/tex] into partial fractions.
[tex]\[ \frac{3 x - 5}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2} \][/tex]
Solving for [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\( 3 x - 5 = A (x-2) + B (x-1) \)[/tex]
Equating coefficients:
[tex]\[ \begin{cases} A + B = 3 \\ -2A - B = -5 \end{cases} \][/tex]
Solving the system of equations, we get [tex]\( A = 2 \)[/tex] and [tex]\( B = 1 \)[/tex].
Therefore:
[tex]\[ \frac{3 x - 5}{(x-1)(x-2)} = \frac{2}{x-1} + \frac{1}{x-2} \][/tex]
3. Integrate each term separately:
[tex]\[ \int \left( \frac{2}{x-1} + \frac{1}{x-2} \right) d x = 2 \int \frac{1}{x-1} d x + \int \frac{1}{x-2} d x \][/tex]
The integrals are:
[tex]\[ 2 \ln|x-1| + \ln|x-2| + C \][/tex]
Combining logarithms:
[tex]\[ \int \frac{3 x - 5}{x^2 - 3 x + 2} d x = \ln| (x-2) | + 2 \ln| (x-1) | + C \][/tex]
Therefore, the result for part (b) is:
[tex]\[ \boxed{\ln|x - 2| + 2 \ln|x - 1| + C} \][/tex]
### (a) [tex]\(\int \frac{d x}{\sqrt{x^2-16}}\)[/tex]
1. Identify the integral form:
The integral [tex]\(\int \frac{d x}{\sqrt{x^2-16}}\)[/tex] resembles a standard integral that can be solved using a trigonometric substitution.
2. Trigonometric substitution:
Consider [tex]\( x = 4 \sec(\theta) \)[/tex], then [tex]\( dx = 4 \sec(\theta) \tan(\theta) d\theta \)[/tex].
3. Substitute and simplify:
Substitute [tex]\( x = 4 \sec(\theta) \)[/tex]:
[tex]\[ \sqrt{x^2 - 16} = \sqrt{16 \sec^2(\theta) - 16} = \sqrt{16 (\sec^2(\theta) - 1)} = 4 \tan(\theta) \][/tex]
So, the integral becomes:
[tex]\[ \int \frac{4 \sec(\theta) \tan(\theta) d\theta}{4 \tan(\theta)} = \int \sec(\theta) d\theta \][/tex]
4. Integrate:
The integral of [tex]\(\sec(\theta)\)[/tex] is [tex]\(\ln|\sec(\theta) + \tan(\theta)| + C\)[/tex].
[tex]\[ \int \sec(\theta) d\theta = \ln|\sec(\theta) + \tan(\theta)| + C \][/tex]
5. Back-substitute [tex]\( \theta \)[/tex]:
Since [tex]\( \sec(\theta) = \frac{x}{4} \)[/tex] and [tex]\( \tan(\theta) = \sqrt{\sec^2(\theta) - 1} = \sqrt{\left(\frac{x}{4}\right)^2 - 1} = \frac{\sqrt{x^2 - 16}}{4} \)[/tex],
[tex]\[ \sec(\theta) + \tan(\theta) = \frac{x}{4} + \frac{\sqrt{x^2 - 16}}{4} = \frac{x + \sqrt{x^2 - 16}}{4} \][/tex]
So, the integral becomes:
[tex]\[ \int \frac{d x}{\sqrt{x^2-16}} = \ln\left| \frac{x + \sqrt{x^2 - 16}}{4} \right| + C \][/tex]
For simplicity, we can remove the constant division inside the logarithm term:
[tex]\[ \int \frac{d x}{\sqrt{x^2-16}} = \ln|x + \sqrt{x^2 - 16}| + C \][/tex]
Therefore, the result for part (a) is:
[tex]\[ \boxed{\ln|x + \sqrt{x^2 - 16}| + C} \][/tex]
### (b) [tex]\(\int \frac{3 x - 5}{x^2 - 3 x + 2} d x\)[/tex]
1. Factor the denominator:
The denominator [tex]\(x^2 - 3x + 2\)[/tex] can be factored as [tex]\((x-1)(x-2)\)[/tex].
[tex]\[ \int \frac{3 x - 5}{(x-1)(x-2)} d x \][/tex]
2. Partial fraction decomposition:
We decompose [tex]\(\frac{3 x - 5}{(x-1)(x-2)}\)[/tex] into partial fractions.
[tex]\[ \frac{3 x - 5}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2} \][/tex]
Solving for [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\( 3 x - 5 = A (x-2) + B (x-1) \)[/tex]
Equating coefficients:
[tex]\[ \begin{cases} A + B = 3 \\ -2A - B = -5 \end{cases} \][/tex]
Solving the system of equations, we get [tex]\( A = 2 \)[/tex] and [tex]\( B = 1 \)[/tex].
Therefore:
[tex]\[ \frac{3 x - 5}{(x-1)(x-2)} = \frac{2}{x-1} + \frac{1}{x-2} \][/tex]
3. Integrate each term separately:
[tex]\[ \int \left( \frac{2}{x-1} + \frac{1}{x-2} \right) d x = 2 \int \frac{1}{x-1} d x + \int \frac{1}{x-2} d x \][/tex]
The integrals are:
[tex]\[ 2 \ln|x-1| + \ln|x-2| + C \][/tex]
Combining logarithms:
[tex]\[ \int \frac{3 x - 5}{x^2 - 3 x + 2} d x = \ln| (x-2) | + 2 \ln| (x-1) | + C \][/tex]
Therefore, the result for part (b) is:
[tex]\[ \boxed{\ln|x - 2| + 2 \ln|x - 1| + C} \][/tex]
