Let's solve each question step-by-step: --- ### Q1: Evaluate the following integrals #### (i) \(\int \cosh x \, e^{\sinh x} \, dx\) **Step 1:** Let \( u = \sinh x \implies du = \cosh x \, dx \). **Step 2:** Substitute into the integral: \[ \int \cosh x \, e^{\sinh x} \, dx = \int e^{\sinh x} \cosh x \, dx = \int e^{u} du \] **Step 3:** Integrate: \[ \int e^{u} du = e^{u} + C \] **Step 4:** Substitute back \( u = \sinh x \): \[ \boxed{e^{\sinh x} + C} \] --- #### (ii) \(\int \frac{\tan^{-1} x}{2 + 2x^2} dx \) **Step 1:** Factor out the 2 from the denominator: \[ \int \frac{\tan^{-1} x}{2(1 + x^2)} dx = \frac{1}{2} \int \frac{\tan^{-1} x}{1 + x^2} dx \] **Step 2:** Let \( u = \tan^{-1} x \implies du = \frac{1}{1 + x^2} dx \) **Step 3:** Substitute: \[ \frac{1}{2} \int u \, du = \frac{1}{2} \cdot \frac{u^2}{2} = \frac{u^2}{4} \] **Step 4:** Substitute back \( u = \tan^{-1} x \): \[ \boxed{\frac{1}{4} (\tan^{-1} x)^2 + C} \] --- ### Q2: Evaluate the following integrals #### (i) \(\int x^2 \sinh x \, dx\) **Step 1:** Use integration by parts. Let \( u = x^2 \), \( dv = \sinh x dx \). Then \( du = 2x dx \), \( v = \cosh x \). By parts: \[ \int u \, dv = uv - \int v \, du \] \[ = x^2 \cosh x - \int 2x \cosh x dx \] **Step 2:** Use integration by parts again for \(\int 2x \cosh x dx\): Let \( u = x \), \( dv = \cosh x dx \implies du = dx \), \( v = \sinh x \) \[ \int x \cosh x dx = x \sinh x - \int \sinh x dx = x \sinh x - \cosh x \] So, \[ \int 2x \cosh x dx = 2(x \sinh x - \cosh x) \] **Step 3:** Substitute back: \[ x^2 \cosh x - 2[x \sinh x - \cosh x] = x^2 \cosh x - 2x \sinh x + 2 \cosh x + C \] So the answer is: \[ \boxed{x^2 \cosh x - 2x \sinh x + 2 \cosh x + C} \] --- #### (ii) \(\int 2x \sin^7(x^2) dx\) **Step 1:** Let \( u = x^2 \implies du = 2x dx \) **Step 2:** Substitute: \[ \int 2x \sin^7(x^2) dx = \int \sin^7(u) du \] **Step 3:** Integrate: \[ \int \sin^7 u \, du \] This can be left as: \[ \boxed{\int \sin^7(x^2) d(x^2) + C} \] Or, for completeness, you can write: \[ \int \sin^7(u) du = -\frac{1}{7} \cos u [1 + 2 \cos^2 u + 3 \cos^4 u] \] But usually, leaving it in terms of the integral is acceptable. --- ### Q3: Evaluate the following integrals #### (i) \(\int \frac{x^2 + 2x}{(x-1)(x^2+1)} dx\) **Step 1:** Partial fraction decomposition: \[ \frac{x^2 + 2x}{(x - 1)(x^2 + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1} \] Multiply both sides by \((x-1)(x^2+1)\): \[ x^2 + 2x = A(x^2 + 1) + (Bx + C)(x-1) \] Expand: \[ x^2 + 2x = A x^2 + A + Bx(x-1) + C(x-1) = A x^2 + A + Bx^2 - Bx + Cx - C = (A + B)x^2 + (C - B)x + (A - C) \] Set coefficients: - \(x^2\): \(A + B = 1\) - \(x\): \(C - B = 2\) - Constant: \(A - C = 0\) Solve: - \(A + B = 1\) - \(C - B = 2\) - \(A = C\) Substitute \(C = A\) into equation 2: \(A - B = 2\) Now two equations: 1. \(A + B = 1\) 2. \(A - B = 2\) Add: \(2A = 3 \implies A = \frac{3}{2}\) Then \(B = 1 - A = 1 - \frac{3}{2} = -\frac{1}{2}\) Then \(C = A = \frac{3}{2}\) So: \[ \frac{x^2 + 2x}{(x-1)(x^2+1)} = \frac{3/2}{x-1} + \frac{ -\frac{1}{2} x + \frac{3}{2} }{x^2 + 1} \] Now integrate: \[ \int \frac{3/2}{x - 1} dx + \int \frac{-\frac{1}{2} x + \frac{3}{2}}{x^2 + 1} dx \] \[ = \frac{3}{2} \ln|x - 1| + \int \frac{-\frac{1}{2} x}{x^2 + 1} dx + \int \frac{3/2}{x^2 + 1} dx \] \[ = \frac{3}{2} \ln|x - 1| - \frac{1}{2} \int \frac{x}{x^2 + 1} dx + \frac{3}{2} \int \frac{1}{x^2 + 1} dx \] \[ \int \frac{x}{x^2 + 1} dx = \frac{1}{2} \ln(x^2 + 1) \] \[ \int \frac{1}{x^2 + 1} dx = \tan^{-1} x \] So final answer: \[ \boxed{ \frac{3}{2} \ln|x - 1| - \frac{1}{4} \ln(x^2 + 1) + \frac{3}{2} \tan^{-1} x + C } \] --- #### (ii) \(\int \frac{x^2 + 3x + 27}{x^3 + 16x} dx\) Factor denominator: \[ x^3 + 16x = x(x^2 + 16) \] Partial fractions: \[ \frac{x^2 + 3x + 27}{x(x^2 + 16)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 16} \] Multiply both sides by \(x(x^2+16)\): \[ x^2 + 3x + 27 = A(x^2 + 16) + (Bx + C)x = A x^2 + 16A + B x^2 + C x = (A + B)x^2 + C x + 16A \] Wait, expand fully: \( (Bx + C)x = Bx^2 + Cx \) So, \[ x^2 + 3x + 27 = A(x^2 + 16) + (Bx + C)x = A x^2 + 16A + B x^2 + C x = (A + B)x^2 + C x + 16A \] Now match coefficients: - \(x^2\): \(A + B = 1\) - \(x\): \(C = 3\) - Constant: \(16A = 27\) So, - \(A = \frac{27}{16}\) - \(B = 1 - A = 1 - \frac{27}{16} = -\frac{11}{16}\) - \(C = 3\) So, \[ \frac{x^2 + 3x + 27}{x(x^2 + 16)} = \frac{27/16}{x} + \frac{ -\frac{11}{16} x + 3 }{ x^2 + 16 } \] Integrate: \[ \int \frac{27/16}{x} dx + \int \frac{ -\frac{11}{16} x + 3 }{ x^2 + 16 } dx \] \[ = \frac{27}{16} \ln|x| - \frac{11}{16} \int \frac{x}{x^2 + 16} dx + 3 \int \frac{1}{x^2 + 16} dx \] \[ \int \frac{x}{x^2 + 16} dx = \frac{1}{2} \ln(x^2 + 16) \] \[ \int \frac{1}{x^2 + 16} dx = \frac{1}{4} \tan^{-1} \left( \frac{x}{4} \right ) \] So, \[ = \frac{27}{16} \ln|x| - \frac{11}{16} \cdot \frac{1}{2} \ln(x^2 + 16) + 3 \cdot \frac{1}{4} \tan^{-1} \left( \frac{x}{4} \right ) \] \[ = \frac{27}{16} \ln|x| - \frac{11}{32} \ln(x^2 + 16) + \frac{3}{4} \tan^{-1} \left( \frac{x}{4} \right ) + C \] \[ \boxed{ \frac{27}{16} \ln|x| - \frac{11}{32} \ln(x^2 + 16) + \frac{3}{4} \tan^{-1} \left( \frac{x}{4} \right ) + C } \] --- ### Q4: Evaluate the following integral \[ \int \frac{1}{\sqrt{2x^2 - 4x + 5}} dx \] Complete the square in the denominator: \(2x^2 - 4x + 5 = 2(x^2 - 2x) + 5\) \(x^2 - 2x = (x - 1)^2 - 1\) So, \(2(x^2 - 2x) = 2[(x - 1)^2 - 1] = 2(x - 1)^2 - 2\) So, \(2x^2 - 4x + 5 = 2(x - 1)^2 - 2 + 5 = 2(x - 1)^2 + 3\) So, \[ \int \frac{1}{\sqrt{2(x - 1)^2 + 3}} dx \] Let \(u = x - 1\), \(dx = du\): \[ \int \frac{1}{\sqrt{2u^2 + 3}} du \] Let’s factor out the 2: \[ = \int \frac{1}{\sqrt{2(u^2 + \frac{3}{2})}} du = \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{u^2 + \frac{3}{2}}} du \] Recall: \[ \int \frac{1}{\sqrt{a^2 + x^2}} dx = \ln \left| x + \sqrt{x^2 + a^2} \right| + C \] Here \(a^2 = \frac{3}{2}\), \(a = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}\) So, \[ = \frac{1}{\sqrt{2}} \ln \left| u + \sqrt{u^2 + \frac{3}{2}} \right| + C

Certainly! Continuing from where I left off: --- Since \( u = x - 1 \), the integral becomes: \[ \int \frac{1}{\sqrt{2u^2 + 3}} du = \frac{1}{\sqrt{2}} \ln \left| u + \sqrt{u^2 + \frac{3}{2}} \right| + C \] Substitute back \( u = x - 1 \): \[ \boxed{ \frac{1}{\sqrt{2}} \ln \left| x - 1 + \sqrt{(x - 1)^2 + \frac{3}{2}} \right| + C } \] --- ### **Final summary of solutions:** 1. \(\int \cosh x\, e^{\sinh x} dx = e^{\sinh x} + C\) 2. \(\int \frac{\tan^{-1} x}{2 + 2x^2} dx = \frac{1}{4} (\tan^{-1} x)^2 + C\) 3. \(\int x^2 \sinh x\, dx = x^2 \cosh x - 2x \sinh x + 2 \cosh x + C\) 4. \(\int 2x \sin^7(x^2) dx = \int \sin^7 u\, du \text{ with } u = x^2\) 5. \(\int \frac{x^2 + 2x}{(x-1)(x^2+1)} dx = \frac{3}{2} \ln|x - 1| - \frac{1}{4} \ln(x^2 + 1) + \frac{3}{2} \tan^{-1} x + C\) 6. \(\int \frac{x^2 + 3x + 27}{x^3 + 16x} dx = \frac{27}{16} \ln|x| - \frac{11}{32} \ln(x^2 + 16) + \frac{3}{4} \tan^{-1} \left( \frac{x}{4} \right) + C\) 7. \(\int \frac{1}{\sqrt{2x^2 - 4x + 5}} dx = \frac{1}{\sqrt{2}} \ln \left| x - 1 + \sqrt{(x - 1)^2 + \frac{3}{2}} \right| + C\) --- **Hope this detailed step-by-step solution helps!**