Let's solve the **first question (1a)** step-by-step: --- ### 1(a) Decompose \( f(x) = \frac{x^2 + 2x + 3}{(x+4)(x-1)(x+1)} \) into partial fractions. #### **Step 1: Set up the partial fraction form** Since the denominator factors are all linear and distinct, write: \[ \frac{x^2 + 2x + 3}{(x+4)(x-1)(x+1)} = \frac{A}{x+4} + \frac{B}{x-1} + \frac{C}{x+1} \] #### **Step 2: Combine into a single fraction** \[ \frac{A}{x+4} + \frac{B}{x-1} + \frac{C}{x+1} = \frac{A(x-1)(x+1) + B(x+4)(x+1) + C(x+4)(x-1)}{(x+4)(x-1)(x+1)} \] #### **Step 3: Equate numerators** \[ x^2 + 2x + 3 = A(x-1)(x+1) + B(x+4)(x+1) + C(x+4)(x-1) \] Expand each term: - \(A(x-1)(x+1) = A(x^2 - 1)\) - \(B(x+4)(x+1) = B(x^2 + 5x + 4)\) - \(C(x+4)(x-1) = C(x^2 + 3x - 4)\) So, \[ x^2 + 2x + 3 = A(x^2 - 1) + B(x^2 + 5x + 4) + C(x^2 + 3x - 4) \] Combine like terms: \[ x^2 + 2x + 3 = (A + B + C)x^2 + (5B + 3C)x + (-A + 4B - 4C) \] #### **Step 4: Equate coefficients** - \(x^2\) terms: \(1 = A + B + C\) - \(x\) terms: \(2 = 5B + 3C\) - Constants: \(3 = -A + 4B - 4C\) So the system is: \[ \begin{cases} A + B + C = 1 \\ 5B + 3C = 2 \\ -A + 4B - 4C = 3 \end{cases} \] #### **Step 5: Solve the system** From the first equation: \(A = 1 - B - C\) Substitute \(A\) into the third equation: \[ -(1 - B - C) + 4B - 4C = 3 \\ -1 + B + C + 4B - 4C = 3 \\ 5B - 3C = 4 \] Now, use the second equation: \[ 5B + 3C = 2 \\ 5B - 3C = 4 \] Add: \[ (5B + 3C) + (5B - 3C) = 2 + 4 \\ 10B = 6 \implies B = \frac{3}{5} \] Substitute \(B\) into one equation, say \(5B + 3C = 2\): \[ 5 \times \frac{3}{5} + 3C = 2 \\ 3 + 3C = 2 \\ 3C = -1 \implies C = -\frac{1}{3} \] Now get \(A\): \[ A = 1 - B - C = 1 - \frac{3}{5} - \left(-\frac{1}{3}\right) = 1 - \frac{3}{5} + \frac{1}{3} \] Find a common denominator (15): \[ 1 = \frac{15}{15}, \quad \frac{3}{5} = \frac{9}{15}, \quad \frac{1}{3} = \frac{5}{15} \] So, \[ A = \frac{15}{15} - \frac{9}{15} + \frac{5}{15} = \frac{11}{15} \] --- ### **Final Partial Fraction Decomposition** \[ \boxed{ \frac{x^2 + 2x + 3}{(x+4)(x-1)(x+1)} = \frac{11}{15(x+4)} + \frac{3}{5(x-1)} - \frac{1}{3(x+1)} } \] --- If you want solutions for the other parts, let me know!

Certainly! Let's proceed systematically through each part of the question, providing detailed solutions, calculations, and final answers. --- ## **Question 1: Decompose each function into partial fractions** ### **Part (a):** Given: \[ f(x) = \frac{x^2 + 2x + 3}{(x+4)(x-1)(x+1)} \] --- ### **Solution:** **Step 1:** Set up partial fractions: \[ \frac{x^2 + 2x + 3}{(x+4)(x-1)(x+1)} = \frac{A}{x+4} + \frac{B}{x-1} + \frac{C}{x+1} \] **Step 2:** Combine RHS over common denominator: \[ A(x-1)(x+1) + B(x+4)(x+1) + C(x+4)(x-1) \] **Step 3:** Expand numerator: - \(A(x-1)(x+1) = A(x^2 - 1)\) - \(B(x+4)(x+1) = B(x^2 + 5x + 4)\) - \(C(x+4)(x-1) = C(x^2 + 3x - 4)\) Total numerator: \[ A(x^2 - 1) + B(x^2 + 5x + 4) + C(x^2 + 3x - 4) \] Combine like terms: \[ (A + B + C) x^2 + (5B + 3C) x + (-A + 4B - 4C) \] Set equal to numerator \(x^2 + 2x + 3\): \[ x^2 + 2x + 3 = (A + B + C) x^2 + (5B + 3C) x + (-A + 4B - 4C) \] **Step 4:** Equate coefficients: \[ \begin{cases} A + B + C = 1 \quad \quad \quad \quad (1) \\ 5B + 3C = 2 \quad \quad \quad \quad (2) \\ -A + 4B - 4C = 3 \quad \quad (3) \end{cases} \] **Step 5:** Solve for \(A, B, C\): From (1): \[ A = 1 - B - C \] Substitute into (3): \[ -(1 - B - C) + 4B - 4C = 3 \] \[ -1 + B + C + 4B - 4C = 3 \] \[ -1 + 5B - 3C = 3 \] \[ 5B - 3C = 4 \quad \quad \quad \quad (4) \] Now, from (2): \[ 5B + 3C = 2 \] Adding (2) and (4): \[ (5B + 3C) + (5B - 3C) = 2 + 4 \] \[ 10B = 6 \] \[ B = \frac{6}{10} = \frac{3}{5} \] Using \(B = \frac{3}{5}\) in (2): \[ 5 \times \frac{3}{5} + 3C = 2 \Rightarrow 3 + 3C = 2 \Rightarrow 3C = -1 \Rightarrow C = -\frac{1}{3} \] Find \(A\): \[ A = 1 - B - C = 1 - \frac{3}{5} - \left(-\frac{1}{3}\right) \] Express all with denominator 15: \[ 1 = \frac{15}{15}, \quad \frac{3}{5} = \frac{9}{15}, \quad -\frac{1}{3} = -\frac{5}{15} \] \[ A = \frac{15}{15} - \frac{9}{15} + \frac{5}{15} = \frac{11}{15} \] --- ### **Final partial fractions:** \[ \boxed{ \frac{x^2 + 2x + 3}{(x+4)(x-1)(x+1)} = \frac{11/15}{x+4} + \frac{3/5}{x-1} - \frac{1/3}{x+1} } \] Or written more cleanly: \[ \boxed{ f(x) = \frac{11}{15(x+4)} + \frac{3}{5(x-1)} - \frac{1}{3(x+1)} } \] --- ## **Part (b):** ### **Given:** \[ f(r) = e^{r} \sin r \] ### **Goal:** Decompose into partial fractions? Actually, this is a function to be integrated, so likely the question wants the integral. --- ### **Part (b):** **Integrate \(f(r) = e^{r} \sin r\) with respect to \(r\).** --- ### **Solution:** **Method:** Use integration by parts or recognize a standard integral involving \(e^{r}\sin r\). **Standard formula:** \[ \int e^{r} \sin r \, dr \] which can be solved using the method of undetermined coefficients or by solving the differential equation approach. --- ### **Step 1:** Set \[ I = \int e^{r} \sin r \, dr \] **Step 2:** Use integration by parts twice: - First, let: \[ u = \sin r \Rightarrow du = \cos r \, dr \] \[ dv = e^{r} dr \Rightarrow v = e^{r} \] - Then: \[ I = e^{r} \sin r - \int e^{r} \cos r \, dr \] Let: \[ J = \int e^{r} \cos r \, dr \] Now, compute \(J\): - Again, use parts: \[ u = \cos r \Rightarrow du = -\sin r \, dr \] \[ dv = e^{r} dr \Rightarrow v = e^{r} \] So: \[ J = e^{r} \cos r + \int e^{r} \sin r \, dr = e^{r} \cos r + I \] **Step 3:** Recall: \[ I = e^{r} \sin r - J \] \[ J = e^{r} \cos r + I \] Substitute \(J\): \[ I = e^{r} \sin r - (e^{r} \cos r + I) \] \[ I = e^{r} \sin r - e^{r} \cos r - I \] \[ I + I = e^{r} \sin r - e^{r} \cos r \] \[ 2I = e^{r} (\sin r - \cos r) \] \[ I = \frac{e^{r}}{2} (\sin r - \cos r) + C \] --- ### **Final answer:** \[ \boxed{ \int e^{r} \sin r \, dr = \frac{e^{r}}{2} (\sin r - \cos r) + C } \] --- ## **Part (c):** ### **Given:** \[ h(z) = 2^{z} \] ### **Goal:** Integrate \(h(z)\) with respect to \(z\). --- ### **Solution:** Recall that: \[ \int a^{z} dz = \frac{a^{z}}{\ln a} + C, \quad a > 0, a \neq 1 \] Here, \(a = 2\): \[ \int 2^{z} dz = \frac{2^{z}}{\ln 2} + C \] --- ### **Final answer:** \[ \boxed{ \int 2^{z} dz = \frac{2^{z}}{\ln 2} + C } \] --- ## **Part 3: Decompose and then integrate** ### **Part (a):** \[ J(t) = \frac{1}{t^2 + 1} \] --- ### **Part (a):** **Decompose \(J(t)\) into partial fractions** Since denominator is quadratic and irreducible, partial fractions may not be applicable unless numerator involves factors. Alternatively, the integral is straightforward: \[ \int \frac{1}{t^2 + 1} dt \] which is a standard integral: \[ \int \frac{1}{t^2 + 1} dt = \arctan t + C \] --- ### **Final answer:** \[ \boxed{ \int \frac{1}{t^2 + 1} dt = \arctan t + C } \] --- ### **Part (b):** \[ K(t) = \frac{1}{t^2 - 1} \] --- ### **Solution:** Factor denominator: \[ t^2 - 1 = (t - 1)(t + 1) \] Set partial fractions: \[ \frac{1}{t^2 - 1} = \frac{A}{t - 1} + \frac{B}{t + 1} \] Multiply both sides by denominator: \[ 1 = A(t + 1) + B(t - 1) \] Set \(t\) values: - For \(t=1\): \[ 1 = A(2) + B(0) \Rightarrow A = \frac{1}{2} \] - For \(t=-1\): \[ 1 = A(0) + B(-2) \Rightarrow B = -\frac{1}{2} \] **Integral:** \[ \int \left( \frac{1/2}{t - 1} - \frac{1/2}{t + 1} \right) dt \] \[ = \frac{1}{2} \int \frac{1}{t - 1} dt - \frac{1}{2} \int \frac{1}{t + 1} dt \] \[ = \frac{1}{2} \ln |t - 1| - \frac{1}{2} \ln |t + 1| + C \] which simplifies to: \[ \boxed{ \frac{1}{2} \ln \left| \frac{t - 1}{t + 1} \right| + C } \] --- ## **Summary of Final Answers:** | Part | Function | Decomposition / Integral | Final Expression | |---------|------------------------------|--------------------------------------------------|-----------------------------------------------------------| | 1(a) | \(\frac{x^2 + 2x + 3}{(x+4)(x-1)(x+1)}\) | Partial fractions | \(\frac{11}{15(x+4)} + \frac{3}{5(x-1)} - \frac{1}{3(x+1)}\) | | 1(b) | \(e^{r} \sin r\) | Integral | \(\frac{e^{r}}{2} (\sin r - \cos r) + C\) | | 1(c) | \(2^{z}\) | Integral | \(\frac{2^{z}}{\ln 2} + C\) | | 3(a) | \(\frac{1}{t^2 + 1}\) | Integral | \(\arctan t + C\) | | 3(b) | \(\frac{1}{t^2 - 1}\) | Partial fractions & integral | \(\frac{1}{2} \ln \left| \frac{t - 1}{t + 1} \right| + C\) | --- If you need further explanation or solutions for other parts, please let me know!

Certainly! Let's proceed with detailed solutions for **Question 2** and its sub-parts. --- ## **Question 2: Integrate each of the following functions with respect to \(x\)** ### **Part (a):** \(\displaystyle f(x) = c^{x} \sin x\) --- ### **Solution:** **Goal:** Find \(\displaystyle \int c^{x} \sin x \, dx\) --- ### **Step 1: Recognize the integral** This is similar to the integral \(\int e^{ax} \sin bx \, dx\), but with base \(c\) instead of \(e\). Recall that: \[ \int e^{ax} \sin bx \, dx = \frac{e^{ax}}{a^2 + b^2} (a \sin bx - b \cos bx) + C \] In our case, \(c^{x} = e^{x \ln c}\), so: \[ \int c^{x} \sin x \, dx = \int e^{x \ln c} \sin x \, dx \] Set: \[ a = \ln c, \quad b = 1 \] --- ### **Step 2: Use the standard integral** \[ \boxed{ \int e^{ax} \sin bx \, dx = \frac{e^{ax}}{a^2 + b^2} (a \sin bx - b \cos bx) + C } \] Applying this: \[ \int c^{x} \sin x \, dx = \frac{e^{x \ln c}}{(\ln c)^2 + 1} \left( \ln c \sin x - \cos x \right) + C \] Recall \(e^{x \ln c} = c^{x}\): \[ \boxed{ \int c^{x} \sin x \, dx = \frac{c^{x}}{(\ln c)^2 + 1} \left( \ln c \sin x - \cos x \right) + C } \] --- ### **Final answer for (a):** \[ \boxed{ \int c^{x} \sin x \, dx = \frac{c^{x}}{(\ln c)^2 + 1} \left( \ln c \sin x - \cos x \right) + C } \] --- ### **Part (b):** \(\displaystyle f(x) = \sin z \cos x\) --- **Note:** The variable \(z\) appears to be independent of \(x\), or perhaps it's a typo. Assuming the function is \(f(x) = \sin x \cos x\). --- ### **Solution:** \[ \int \sin x \cos x \, dx \] This is a standard integral; use the double-angle identity: \[ \sin x \cos x = \frac{1}{2} \sin 2x \] So, \[ \int \sin x \cos x \, dx = \frac{1}{2} \int \sin 2x \, dx \] \[ = \frac{1}{2} \left( - \frac{\cos 2x}{2} \right) + C = - \frac{1}{4} \cos 2x + C \] --- ### **Final answer for (b):** \[ \boxed{ \int \sin x \cos x \, dx = - \frac{1}{4} \cos 2x + C } \] --- ### **Part (c):** \(\displaystyle h(x) = 2 \times 10^{x}\) --- ### **Solution:** Recall that: \[ \int a^{x} dx = \frac{a^{x}}{\ln a} + C \] Here, \(a = 10\), and the coefficient 2 is constant. \[ \int 2 \times 10^{x} dx = 2 \int 10^{x} dx = 2 \times \frac{10^{x}}{\ln 10} + C \] --- ### **Final answer for (c):** \[ \boxed{ \int 2 \times 10^{x} dx = \frac{2 \times 10^{x}}{\ln 10} + C } \] --- ## **Summary of Question 2 answers:** | Part | Function | Integral | Final Expression | |---------|------------------------------|----------------------------------------------------------|--------------------------------------------------------------| | (a) | \(c^{x} \sin x\) | \(\int c^{x} \sin x \, dx\) | \(\frac{c^{x}}{(\ln c)^2 + 1} (\ln c \sin x - \cos x) + C\) | | (b) | \(\sin x \cos x\) | \(\int \sin x \cos x \, dx\) | \(- \frac{1}{4} \cos 2x + C\) | | (c) | \(2 \times 10^{x}\) | \(\int 2 \times 10^{x} dx\) | \(\frac{2 \times 10^{x}}{\ln 10} + C\) | --- If you'd like me to clarify further or proceed with other parts, please let me know!

Certainly! Here are the final answers for all parts of the questions: --- ### **Question 1: Partial Fraction Decomposition and Integrations** **1(a):** \(\displaystyle \frac{x^2 + 2x + 3}{(x+4)(x-1)(x+1)} = \frac{11}{15(x+4)} + \frac{3}{5(x-1)} - \frac{1}{3(x+1)}\) **1(b):** \(\displaystyle \int e^{r} \sin r \, dr = \frac{e^{r}}{2} (\sin r - \cos r) + C\) **1(c):** \(\displaystyle \int 2^{z} \, dz = \frac{2^{z}}{\ln 2} + C\) --- ### **Question 2: Integrations** **2(a):** \(\displaystyle \int c^{x} \sin x \, dx = \frac{c^{x}}{(\ln c)^2 + 1} \left( \ln c \sin x - \cos x \right) + C\) **2(b):** \(\displaystyle \int \sin x \cos x \, dx = -\frac{1}{4} \cos 2x + C\) **2(c):** \(\displaystyle \int 2 \times 10^{x} \, dx = \frac{2 \times 10^{x}}{\ln 10} + C\) --- ### **Question 3: Decompose and Integrate** **3(a):** \(\displaystyle \int \frac{1}{t^2 + 1} \, dt = \arctan t + C\) **3(b):** \(\displaystyle \int \frac{1}{t^2 - 1} \, dt = \frac{1}{2} \ln \left| \frac{t - 1}{t + 1} \right| + C\) --- Let me know if you need further assistance!