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Question 1. Integrate: \(\displaystyle \int \frac{4+5x^{3/2}}{\sqrt{x}}\,dx\).
Simplify the integrand: \(\displaystyle \frac{4+5x^{3/2}}{\sqrt{x}}=4x^{-1/2}+5x\). Integrate term by term: \(\displaystyle \int4x^{-1/2}dx=8x^{1/2}=8\sqrt{x}\), and \(\displaystyle \int5x\,dx=\frac52x^2\). Therefore the integral is \(\displaystyle 8\sqrt{x}+\frac52x^2+C\). Answer: B.
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Question 2. Integrate: \(\displaystyle \int \sqrt[3]{x^2}\,dx\).
Rewrite \(\sqrt[3]{x^2}\) as \(x^{2/3}\). Using the power rule, \(\displaystyle \int x^{2/3}dx=\frac{x^{5/3}}{5/3}+C=\frac35x^{5/3}+C\). Answer: B.
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Question 3. Find the particular solution of the equation \(f'(x)=4x^{-1/2}\) that satisfies \(f(1)=12\).
Integrate \(f'(x)=4x^{-1/2}\): \(\displaystyle f(x)=4\cdot\frac{x^{1/2}}{1/2}+C=8\sqrt{x}+C\). Use \(f(1)=12\): \(8+C=12\), so \(C=4\). Thus \(f(x)=8\sqrt{x}+4\). Answer: D.
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Question 4. Find \(y=f(x)\) if \(f''(x)=x+2\), \(f'(0)=3\), and \(f(0)=-1\).
Integrate \(f''(x)=x+2\): \(\displaystyle f'(x)=\frac12x^2+2x+C_1\). Using \(f'(0)=3\), \(C_1=3\). Integrate again: \(\displaystyle f(x)=\frac16x^3+x^2+3x+C_2\). Using \(f(0)=-1\), \(C_2=-1\). Therefore \(f(x)=\frac16x^3+x^2+3x-1\). Answer: A.
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Question 5. Integrate: \(\displaystyle \int 5\sec x\tan x\,dx\).
Since \(\displaystyle \frac{d}{dx}(\sec x)=\sec x\tan x\), \(\displaystyle \int5\sec x\tan x\,dx=5\sec x+C\). Answer: B.
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Question 6. Evaluate: \(\displaystyle \int 10^x\,dx\).
Use \(\displaystyle \int a^x\,dx=\frac{a^x}{\ln a}+C\). With \(a=10\), \(\displaystyle \int10^x\,dx=\frac{10^x}{\ln10}+C\). Answer: C.
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Question 7. Evaluate: \(\displaystyle \int\frac{(e^x+e^{2x})^2}{e^{3x}}\,dx\).
Expand and simplify: \((e^x+e^{2x})^2=e^{2x}+2e^{3x}+e^{4x}\). Dividing by \(e^{3x}\) gives \(e^{-x}+2+e^x\). Integrate term by term: \(\displaystyle \int e^{-x}dx=-e^{-x}\), \(\int2dx=2x\), and \(\int e^xdx=e^x\). Thus the result is \(e^x-e^{-x}+2x+C\). Answer: C.
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Question 8. Evaluate the indefinite integral: \(\displaystyle \int(\ln x)^4\,dx\).
Use integration by parts with \(u=(\ln x)^4\) and \(dv=dx\). Then \(du=4(\ln x)^3\frac1x\,dx\) and \(v=x\). Hence \(\displaystyle \int(\ln x)^4dx=x(\ln x)^4-4\int(\ln x)^3dx\). Answer: A.
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Question 9. Evaluate: \(\displaystyle \int\sin(\ln x)\,dx\).
Let \(t=\ln x\), so \(x=e^t\) and \(dx=e^tdt\). Then \(\displaystyle \int\sin(\ln x)dx=\int e^t\sin t\,dt\). Using integration by parts twice, \(\displaystyle \int e^t\sin t\,dt=\frac{e^t}{2}(\sin t-\cos t)+C\). Substitute back to get \(\displaystyle \frac{x\sin(\ln x)}2-\frac{x\cos(\ln x)}2+C\). Answer: B.
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Question 10. Integrate: \(\displaystyle \int\frac{\sec^3\theta\tan\theta}{1+\tan^2\theta}\,d\theta\).
Use the identity \(1+\tan^2\theta=\sec^2\theta\). The integrand simplifies to \(\displaystyle \frac{\sec^3\theta\tan\theta}{\sec^2\theta}=\sec\theta\tan\theta\). Since \(\frac{d}{d\theta}(\sec\theta)=\sec\theta\tan\theta\), the integral is \(\sec\theta+C\). Answer: D.
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Question 11. Integrate: \(\displaystyle \int\frac1{x^3}\,dx\).
Rewrite \(\frac1{x^3}\) as \(x^{-3}\). By the power rule, \(\displaystyle \int x^{-3}dx=\frac{x^{-2}}{-2}+C=-\frac1{2x^2}+C\). Answer: A.
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Question 12. Integrate: \(\displaystyle \int(4x^4-2x^2+3)\,dx\).
Integrate each term separately: \(\int4x^4dx=\frac45x^5\), \(\int-2x^2dx=-\frac23x^3\), and \(\int3dx=3x\). Therefore the result is \(\displaystyle \frac45x^5-\frac23x^3+3x+C\). Answer: E.
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Question 13. Integrate: \(\displaystyle \int(4x^3-2x+3)\,dx\).
Apply the power rule term by term: \(\int4x^3dx=x^4\), \(\int-2x\,dx=-x^2\), and \(\int3dx=3x\). Thus the integral is \(x^4-x^2+3x+C\). Answer: D.
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Question 14. Find an expression in \(a\) and \(b\) for \(\displaystyle \int_a^b\left(5x^2+\sqrt{x-7}\right)\,dx\).
An antiderivative is \(\displaystyle F(x)=\frac53x^3+\frac23(x-7)^{3/2}\). Using the Fundamental Theorem of Calculus, \(\displaystyle F(b)-F(a)\) equals \(\displaystyle \frac53(b^3-a^3)+\frac23(b-7)^{3/2}-\frac23(a-7)^{3/2}\). Answer: A.
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Question 15. Evaluate: \(\displaystyle \int\frac{5x+16}{x^2+9}\,dx\).
Split the integral: \(\displaystyle \int\frac{5x}{x^2+9}dx+16\int\frac1{x^2+9}dx\). For the first part, let \(u=x^2+9\), so \(du=2x\,dx\): \(\displaystyle \int\frac{5x}{x^2+9}dx=\frac52\ln(x^2+9)\). For the second part, use \(\displaystyle \int\frac1{x^2+a^2}dx=\frac1a\arctan\left(\frac xa\right)\): \(\displaystyle 16\int\frac1{x^2+9}dx=\frac{16}{3}\arctan\left(\frac x3\right)\). Therefore the result is \(\displaystyle \frac52\ln(x^2+9)+\frac{16}{3}\arctan\left(\frac x3\right)+C\). Answer: A.
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