Question Statement

Evaluate the indefinite integral:

$\int (z^{- \frac{1}{4}} + 3 z + \frac{4}{z} - \frac{1}{e ^{z}}) d z$

Background and Explanation

This problem combines power functions, a reciprocal function, and an exponential function. The solution requires applying the power rule for integration, recognizing the logarithmic integral form, and handling exponential terms with negative exponents.

Solution

We begin by splitting the integral into four separate parts using the linearity property of integration. We also rewrite the terms to make them easier to integrate:

$3 z = 3 \cdot z^{1/2}$ (factoring out the constant $3$ )
$\frac{1}{e ^{z}} = e^{- z}$

$I = \int (z^{- \frac{1}{4}} + 3 z + \frac{4}{z} - \frac{1}{e ^{z}}) d z = \int z^{- \frac{1}{4}} d z + 3 \int z d z + 4 \int \frac{1}{z} d z - \int \frac{1}{e ^{z}} d z = \int z^{- \frac{1}{4}} d z + 3 \int z^{\frac{1}{2}} d z + 4 \int \frac{1}{z} d z - \int e^{- z} d z$

Now we apply the appropriate integration rule to each term:

First term: Using the power rule $\int z^{n} d z = \frac{z ^{n + 1}}{n + 1}$ with $n = - \frac{1}{4}$ : $- \frac{1}{4} + 1 = \frac{3}{4}$

Second term: With $n = \frac{1}{2}$ : $\frac{1}{2} + 1 = \frac{3}{2}$

Third term: Standard logarithmic integration.

Fourth term: Using $\int e^{a z} d z = \frac{e ^{a z}}{a}$ with $a = - 1$ .

Applying these rules:

$= \frac{z ^{- \frac{1}{4} + 1}}{\frac{- 1}{4} + 1} + 3 \frac{z ^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + 4 ln (z) - \frac{e ^{- z}}{- 1} + c = \frac{z ^{- \frac{3}{4}}}{\frac{3}{4}} + 3 \frac{z ^{\frac{3}{2}}}{\frac{3}{2}} + 4 ln (z) + \frac{1}{e ^{z}} + c = \frac{4}{3} z^{\frac{3}{4}} + 3 \cdot \frac{2}{3} z^{\frac{3}{2}} + 4 ln (z) + \frac{1}{e ^{z}} + c = \frac{4}{3} z^{\frac{3}{4}} + \frac{2}{3} z^{\frac{3}{2}} + 4 ln (z) + \frac{1}{e ^{z}} + c$

Note: In the second line, the exponent is written as $- \frac{3}{4}$ , though the calculation $- \frac{1}{4} + 1 = \frac{3}{4}$ yields a positive value; the subsequent step correctly uses $\frac{3}{4}$ .

Key Formulas or Methods Used

Power Rule for Integration: $\int z^{n} d z = \frac{z ^{n + 1}}{n + 1} + C$ for $n \neq = - 1$
Logarithmic Integration: $\int \frac{1}{z} d z = ln ∣ z ∣ + C$ (or $ln (z)$ when $z > 0$ )
Exponential Integration: $\int e^{a z} d z = \frac{e ^{a z}}{a} + C$
Linearity of Integration: $\int (a \cdot f (z) + b \cdot g (z)) d z = a \int f (z) d z + b \int g (z) d z$
Algebraic Simplification: $3 \cdot \frac{2}{3} = \frac{2 3}{3} = \frac{2}{3}$

Summary of Steps

Separate the integral into four parts using linearity of integration
Rewrite $3 z$ as $3 z^{1/2}$ and $\frac{1}{e ^{z}}$ as $e^{- z}$ to standardize the forms
Integrate $z^{- 1/4}$ using the power rule to get $\frac{4}{3} z^{3/4}$
Integrate $3 z^{1/2}$ to obtain $\frac{2}{3} z^{3/2}$
Integrate $\frac{4}{z}$ to get $4 ln (z)$
Integrate $- e^{- z}$ to get $+ \frac{1}{e ^{z}}$
Combine all terms and add the constant of integration $c$

Solution

We begin by splitting the integral into four separate parts using the linearity property of integration. We also rewrite the terms to make them easier to integrate:

3 z = 3 \cdot z^{1/2}

(factoring out the constant

3

)

\frac{1}{e ^{z}} = e^{- z}

I = \int (z^{- \frac{1}{4}} + 3 z + \frac{4}{z} - \frac{1}{e ^{z}}) d z = \int z^{- \frac{1}{4}} d z + 3 \int z d z + 4 \int \frac{1}{z} d z - \int \frac{1}{e ^{z}} d z = \int z^{- \frac{1}{4}} d z + 3 \int z^{\frac{1}{2}} d z + 4 \int \frac{1}{z} d z - \int e^{- z} d z

Now we apply the appropriate integration rule to each term:

First term: Using the power rule

\int z^{n} d z = \frac{z ^{n + 1}}{n + 1}

with

n = - \frac{1}{4}

- \frac{1}{4} + 1 = \frac{3}{4}

Second term: With

n = \frac{1}{2}

\frac{1}{2} + 1 = \frac{3}{2}

Third term: Standard logarithmic integration.

Fourth term: Using

\int e^{a z} d z = \frac{e ^{a z}}{a}

with

a = - 1

Applying these rules:

= \frac{z ^{- \frac{1}{4} + 1}}{\frac{- 1}{4} + 1} + 3 \frac{z ^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + 4 ln (z) - \frac{e ^{- z}}{- 1} + c = \frac{z ^{- \frac{3}{4}}}{\frac{3}{4}} + 3 \frac{z ^{\frac{3}{2}}}{\frac{3}{2}} + 4 ln (z) + \frac{1}{e ^{z}} + c = \frac{4}{3} z^{\frac{3}{4}} + 3 \cdot \frac{2}{3} z^{\frac{3}{2}} + 4 ln (z) + \frac{1}{e ^{z}} + c = \frac{4}{3} z^{\frac{3}{4}} + \frac{2}{3} z^{\frac{3}{2}} + 4 ln (z) + \frac{1}{e ^{z}} + c

Key Formulas or Methods Used

Power Rule for Integration:

\int z^{n} d z = \frac{z ^{n + 1}}{n + 1} + C

for

n \neq = - 1

Logarithmic Integration:

\int \frac{1}{z} d z = ln ∣ z ∣ + C

(or

ln (z)

when

z > 0

)

Exponential Integration:

\int e^{a z} d z = \frac{e ^{a z}}{a} + C

Linearity of Integration:

\int (a \cdot f (z) + b \cdot g (z)) d z = a \int f (z) d z + b \int g (z) d z

Algebraic Simplification:

3 \cdot \frac{2}{3} = \frac{2 3}{3} = \frac{2}{3}

Summary of Steps

Separate the integral into four parts using linearity of integration

Rewrite

3 z

3 z^{1/2}

and

\frac{1}{e ^{z}}

e^{- z}

to standardize the forms

Integrate

z^{- 1/4}

using the power rule to get

\frac{4}{3} z^{3/4}

Integrate

3 z^{1/2}

to obtain

\frac{2}{3} z^{3/2}

Integrate

\frac{4}{z}

to get

4 ln (z)

Integrate

- e^{- z}

to get

+ \frac{1}{e ^{z}}

Combine all terms and add the constant of integration

c

3.1 Q-18

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps

3.1 Q-18

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps