Answer :
Let’s tackle each integral step-by-step to find the general antiderivatives.
### Part a:
Given the integral:
[tex]\[ \int \frac{2 x^3-5 x+x^{3 / 2}}{4 x^3} \, dx \][/tex]
First, simplify the integrand:
[tex]\[ \frac{2 x^3-5 x+x^{3 / 2}}{4 x^3} = \frac{2 x^3}{4 x^3} - \frac{5 x}{4 x^3} + \frac{x^{3/2}}{4 x^3} \][/tex]
[tex]\[ = \frac{1}{2} - \frac{5}{4x^2} + \frac{x^{3/2 - 3}}{4} = \frac{1}{2} - \frac{5}{4x^2} + \frac{1}{4x^{3/2}} \][/tex]
Now, integrate each term separately:
[tex]\[ \int \left(\frac{1}{2} - \frac{5}{4x^2} + \frac{1}{4x^{3/2}}\right) \, dx \][/tex]
### Integral of [tex]\(\frac{1}{2}\)[/tex]:
[tex]\[ \int \frac{1}{2} \, dx = \frac{1}{2} x \][/tex]
### Integral of [tex]\(- \frac{5}{4x^2}\)[/tex]:
Rewrite as:
[tex]\[ \int - \frac{5}{4} x^{-2} \, dx = - \frac{5}{4} \int x^{-2} \, dx \][/tex]
[tex]\[ = - \frac{5}{4} \cdot \left(-x^{-1}\right) = \frac{5}{4x} \][/tex]
### Integral of [tex]\(\frac{1}{4x^{3/2}}\)[/tex]:
Rewrite as:
[tex]\[ \int \frac{1}{4} x^{-3/2} \, dx = \frac{1}{4} \int x^{-3/2} \, dx \][/tex]
[tex]\[ = \frac{1}{4} \cdot \left(\frac{x^{-1/2}}{-1/2}\right) = -\frac{1}{2x^{1/2}} = -\frac{1}{2\sqrt{x}} \][/tex]
Combining all these results, we have:
[tex]\[ \int \frac{2 x^3 - 5 x + x^{3/2}}{4 x^3} \, dx = \frac{1}{2} x + \frac{5}{4x} - \frac{1}{2\sqrt{x}} + C \][/tex]
### Part b:
Given the integral:
[tex]\[ \int \frac{17 \cos x}{1 - \cos^2 x} \, dx \][/tex]
Recall the Pythagorean identity:
[tex]\[ 1 - \cos^2 x = \sin^2 x \][/tex]
Therefore, we can rewrite the integrand as:
[tex]\[ \int \frac{17 \cos x}{\sin^2 x} \, dx = 17 \int \frac{\cos x}{\sin^2 x} \, dx \][/tex]
[tex]\[ = 17 \int \frac{d(\sin x)}{\sin^2 x} \, dx = 17 \int \csc^2 x \, d(\sin x) \][/tex]
The antiderivative of [tex]\(\csc^2 x\)[/tex] is:
[tex]\[ \int \csc^2 x \, dx = -\cot x \][/tex]
Therefore:
[tex]\[ 17 \int \csc^2 x \, dx = -17 \cot x = -17 \frac{\cos x}{\sin x} \][/tex]
So the result is:
[tex]\[ \int \frac{17 \cos x}{1 - \cos^2 x} \, dx = -17 \cot x + C \][/tex]
Alternatively, using the half-angle formula:
[tex]\[ \cot(x) = \frac{\cos(x)}{\sin(x)} = \frac{\cos(x/2)^2 - \sin(x/2)^2}{2 \cos(x/2) \sin(x/2)} = \cot(x/2) - \tan(x/2) \][/tex]
Thus:
[tex]\[ \int \frac{17 \cos x}{1 - \cos^2 x} \, dx = -\frac{17}{2} \left( \tan \left(\frac{x}{2}\right) + \frac{1}{\tan \left(\frac{x}{2}\right)} \right) + C \][/tex]
[tex]\[ = -\frac{17}{2} \tan \left(\frac{x}{2}\right) - \frac{17}{2 \tan \left(\frac{x}{2}\right)} + C \][/tex]
### Part c:
Given the integral:
[tex]\[ \int \left( 2^x - \frac{7}{x^2 + 1} \right) \, dx \][/tex]
### Integral of [tex]\(2^x\)[/tex]:
Recall:
[tex]\[ \int 2^x \, dx = \frac{2^x}{\ln(2)} \][/tex]
### Integral of [tex]\(\frac{7}{x^2 + 1}\)[/tex]:
Recall:
[tex]\[ \int \frac{1}{x^2 + 1} \, dx = \arctan(x) \][/tex]
So:
[tex]\[ \int \frac{7}{x^2 + 1} \, dx = 7 \arctan(x) \][/tex]
Combining these results, we have:
[tex]\[ \int \left( 2^x - \frac{7}{x^2 + 1} \right) \, dx = \frac{2^x}{\ln(2)} - 7 \arctan(x) + C \][/tex]
### Summary:
a. [tex]\[ \int \frac{2 x^3-5 x+x^{3 / 2}}{4 x^3} \, dx = \frac{1}{2} x + \frac{5}{4 x} - \frac{1}{2 \sqrt{x}} + C \][/tex]
b. [tex]\[ \int \frac{17 \cos x}{1-\cos ^2 x} \, dx = -\frac{17}{2} \left(\tan \left(\frac{x}{2}\right) + \frac{1}{\tan \left(\frac{x}{2}\right)}\right) + C \][/tex]
c. [tex]\[ \int \left(2^x - \frac{7}{x^2 + 1}\right) \, dx = \frac{2^x}{\ln(2)} - 7 \arctan(x) + C \][/tex]
Remember to always add the integration constant [tex]\(C\)[/tex] in indefinite integrals!
### Part a:
Given the integral:
[tex]\[ \int \frac{2 x^3-5 x+x^{3 / 2}}{4 x^3} \, dx \][/tex]
First, simplify the integrand:
[tex]\[ \frac{2 x^3-5 x+x^{3 / 2}}{4 x^3} = \frac{2 x^3}{4 x^3} - \frac{5 x}{4 x^3} + \frac{x^{3/2}}{4 x^3} \][/tex]
[tex]\[ = \frac{1}{2} - \frac{5}{4x^2} + \frac{x^{3/2 - 3}}{4} = \frac{1}{2} - \frac{5}{4x^2} + \frac{1}{4x^{3/2}} \][/tex]
Now, integrate each term separately:
[tex]\[ \int \left(\frac{1}{2} - \frac{5}{4x^2} + \frac{1}{4x^{3/2}}\right) \, dx \][/tex]
### Integral of [tex]\(\frac{1}{2}\)[/tex]:
[tex]\[ \int \frac{1}{2} \, dx = \frac{1}{2} x \][/tex]
### Integral of [tex]\(- \frac{5}{4x^2}\)[/tex]:
Rewrite as:
[tex]\[ \int - \frac{5}{4} x^{-2} \, dx = - \frac{5}{4} \int x^{-2} \, dx \][/tex]
[tex]\[ = - \frac{5}{4} \cdot \left(-x^{-1}\right) = \frac{5}{4x} \][/tex]
### Integral of [tex]\(\frac{1}{4x^{3/2}}\)[/tex]:
Rewrite as:
[tex]\[ \int \frac{1}{4} x^{-3/2} \, dx = \frac{1}{4} \int x^{-3/2} \, dx \][/tex]
[tex]\[ = \frac{1}{4} \cdot \left(\frac{x^{-1/2}}{-1/2}\right) = -\frac{1}{2x^{1/2}} = -\frac{1}{2\sqrt{x}} \][/tex]
Combining all these results, we have:
[tex]\[ \int \frac{2 x^3 - 5 x + x^{3/2}}{4 x^3} \, dx = \frac{1}{2} x + \frac{5}{4x} - \frac{1}{2\sqrt{x}} + C \][/tex]
### Part b:
Given the integral:
[tex]\[ \int \frac{17 \cos x}{1 - \cos^2 x} \, dx \][/tex]
Recall the Pythagorean identity:
[tex]\[ 1 - \cos^2 x = \sin^2 x \][/tex]
Therefore, we can rewrite the integrand as:
[tex]\[ \int \frac{17 \cos x}{\sin^2 x} \, dx = 17 \int \frac{\cos x}{\sin^2 x} \, dx \][/tex]
[tex]\[ = 17 \int \frac{d(\sin x)}{\sin^2 x} \, dx = 17 \int \csc^2 x \, d(\sin x) \][/tex]
The antiderivative of [tex]\(\csc^2 x\)[/tex] is:
[tex]\[ \int \csc^2 x \, dx = -\cot x \][/tex]
Therefore:
[tex]\[ 17 \int \csc^2 x \, dx = -17 \cot x = -17 \frac{\cos x}{\sin x} \][/tex]
So the result is:
[tex]\[ \int \frac{17 \cos x}{1 - \cos^2 x} \, dx = -17 \cot x + C \][/tex]
Alternatively, using the half-angle formula:
[tex]\[ \cot(x) = \frac{\cos(x)}{\sin(x)} = \frac{\cos(x/2)^2 - \sin(x/2)^2}{2 \cos(x/2) \sin(x/2)} = \cot(x/2) - \tan(x/2) \][/tex]
Thus:
[tex]\[ \int \frac{17 \cos x}{1 - \cos^2 x} \, dx = -\frac{17}{2} \left( \tan \left(\frac{x}{2}\right) + \frac{1}{\tan \left(\frac{x}{2}\right)} \right) + C \][/tex]
[tex]\[ = -\frac{17}{2} \tan \left(\frac{x}{2}\right) - \frac{17}{2 \tan \left(\frac{x}{2}\right)} + C \][/tex]
### Part c:
Given the integral:
[tex]\[ \int \left( 2^x - \frac{7}{x^2 + 1} \right) \, dx \][/tex]
### Integral of [tex]\(2^x\)[/tex]:
Recall:
[tex]\[ \int 2^x \, dx = \frac{2^x}{\ln(2)} \][/tex]
### Integral of [tex]\(\frac{7}{x^2 + 1}\)[/tex]:
Recall:
[tex]\[ \int \frac{1}{x^2 + 1} \, dx = \arctan(x) \][/tex]
So:
[tex]\[ \int \frac{7}{x^2 + 1} \, dx = 7 \arctan(x) \][/tex]
Combining these results, we have:
[tex]\[ \int \left( 2^x - \frac{7}{x^2 + 1} \right) \, dx = \frac{2^x}{\ln(2)} - 7 \arctan(x) + C \][/tex]
### Summary:
a. [tex]\[ \int \frac{2 x^3-5 x+x^{3 / 2}}{4 x^3} \, dx = \frac{1}{2} x + \frac{5}{4 x} - \frac{1}{2 \sqrt{x}} + C \][/tex]
b. [tex]\[ \int \frac{17 \cos x}{1-\cos ^2 x} \, dx = -\frac{17}{2} \left(\tan \left(\frac{x}{2}\right) + \frac{1}{\tan \left(\frac{x}{2}\right)}\right) + C \][/tex]
c. [tex]\[ \int \left(2^x - \frac{7}{x^2 + 1}\right) \, dx = \frac{2^x}{\ln(2)} - 7 \arctan(x) + C \][/tex]
Remember to always add the integration constant [tex]\(C\)[/tex] in indefinite integrals!
