Question 1. Integrate: \(\displaystyle \int \frac{1}{x^3}\,dx\).
Rewrite the integrand as \(x^{-3}\). Using the power rule, \(\displaystyle \int x^n\,dx=\frac{x^{n+1}}{n+1}+C\) for \(n\ne-1\). Thus \(\displaystyle \int x^{-3}\,dx=\frac{x^{-2}}{-2}+C=-\frac{1}{2x^2}+C\). Answer: A.
Question 2. 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 \(f(1)=12\): \(8\sqrt{1}+C=12\), so \(C=4\). Therefore \(f(x)=8\sqrt{x}+4\). Answer: D.
Question 3. Evaluate: \(\displaystyle \int x\cos(2x)\,dx\).
Use integration by parts with \(u=x\) and \(dv=\cos(2x)\,dx\). Then \(du=dx\) and \(v=\frac12\sin(2x)\). Therefore \(\displaystyle \int x\cos(2x)\,dx=\frac{x}{2}\sin(2x)-\frac12\int\sin(2x)\,dx\). Since \(\displaystyle \int\sin(2x)\,dx=-\frac12\cos(2x)\), the result is \(\displaystyle \frac12x\sin(2x)+\frac14\cos(2x)+C\). Answer: C.
Question 4. Evaluate: \(\displaystyle \int x(x^2-1)^4\,dx\).
Let \(u=x^2-1\). Then \(du=2x\,dx\), so \(x\,dx=\frac12du\). Thus \(\displaystyle \int x(x^2-1)^4\,dx=\frac12\int u^4\,du=\frac12\cdot\frac{u^5}{5}+C\). Substitute back: \(\displaystyle \frac{1}{10}(x^2-1)^5+C\). Answer: B.
Question 5. Integrate: \(\displaystyle \int 5\sec x\tan x\,dx\).
Recall that \(\displaystyle \frac{d}{dx}(\sec x)=\sec x\tan x\). Therefore \(\displaystyle \int5\sec x\tan x\,dx=5\sec x+C\). Answer: B.
Question 6. 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 \(\displaystyle \int(\ln x)^4\,dx=x(\ln x)^4-\int x\cdot4(\ln x)^3\frac1x\,dx\). So \(\displaystyle \int(\ln x)^4\,dx=x(\ln x)^4-4\int(\ln x)^3\,dx\). Answer: A.
Question 7. Evaluate: \(\displaystyle \int x^2\ln x\,dx\).
Use integration by parts with \(u=\ln x\) and \(dv=x^2dx\). Then \(du=\frac1x dx\) and \(v=\frac{x^3}{3}\). Therefore \(\displaystyle \int x^2\ln x\,dx=\frac{x^3}{3}\ln x-\frac13\int x^2\,dx\). This equals \(\displaystyle \frac{x^3\ln x}{3}-\frac{x^3}{9}+C\). Answer: E.
Question 8. Evaluate: \(\displaystyle \int10^x\,dx\).
Use the exponential integral rule \(\displaystyle \int a^x\,dx=\frac{a^x}{\ln a}+C\) for \(a>0\), \(a\ne1\). With \(a=10\), the integral is \(\displaystyle \frac{10^x}{\ln10}+C\). Answer: C.
Question 9. Evaluate: \(\displaystyle \int\cot^2x\,dx\).
Use the identity \(\cot^2x=\csc^2x-1\). Then \(\displaystyle \int\cot^2x\,dx=\int(\csc^2x-1)\,dx\). Since \(\displaystyle \int\csc^2x\,dx=-\cot x\), the result is \(-\cot x-x+C\). Answer: C.
Question 10. Evaluate: \(\displaystyle \int\frac{dx}{(x-2)(x+3)}\).
Use partial fractions: \(\displaystyle \frac{1}{(x-2)(x+3)}=\frac{A}{x-2}+\frac{B}{x+3}\). Solving \(1=A(x+3)+B(x-2)\) gives \(A=\frac15\) and \(B=-\frac15\). Therefore \(\displaystyle \int\frac{dx}{(x-2)(x+3)}=\frac15\ln|x-2|-\frac15\ln|x+3|+C\). Answer: B.
Question 11. Integrate: \(\displaystyle \int(4x^4-2x^2+3)\,dx\).
Integrate term by term using the power rule: \(\displaystyle \int4x^4dx=\frac45x^5\), \(\displaystyle \int-2x^2dx=-\frac23x^3\), and \(\displaystyle \int3dx=3x\). Thus the integral is \(\displaystyle \frac45x^5-\frac23x^3+3x+C\). Answer: E.
Question 12. Evaluate: \(\displaystyle \int\frac{x}{(x-3)^{2/3}}\,dx\).
Let \(u=x-3\), so \(x=u+3\) and \(du=dx\). Then \(\displaystyle \int\frac{x}{(x-3)^{2/3}}dx=\int\frac{u+3}{u^{2/3}}du=\int\left(u^{1/3}+3u^{-2/3}\right)du\). Integrating gives \(\displaystyle \frac34u^{4/3}+9u^{1/3}+C\). Substitute \(u=x-3\): \(\displaystyle \frac34(x-3)^{4/3}+9(x-3)^{1/3}+C\). Answer: D.
Question 13. Evaluate: \(\displaystyle \int\sin(\ln x)\,dx\).
Let \(t=\ln x\). Then \(x=e^t\) and \(dx=e^t dt\), so \(\displaystyle \int\sin(\ln x)\,dx=\int e^t\sin t\,dt\). Using the standard result \(\displaystyle \int e^t\sin t\,dt=\frac{e^t}{2}(\sin t-\cos t)+C\), and substituting \(e^t=x\), \(t=\ln x\), we get \(\displaystyle \frac{x\sin(\ln x)}2-\frac{x\cos(\ln x)}2+C\). Answer: B.
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 \(\displaystyle F(x)=\frac53x^3+\frac23(x-7)^{3/2}\). By the Fundamental Theorem of Calculus, the definite integral is \(\displaystyle F(b)-F(a)\). Therefore the value is \(\displaystyle \frac53(b^3-a^3)+\frac23(b-7)^{3/2}-\frac23(a-7)^{3/2}\). Answer: A.
Question 15. Find \(y=f(x)\) if \(f''(x)=x^2\), \(f'(0)=7\), and \(f(0)=2\).
Integrate \(f''(x)=x^2\): \(\displaystyle f'(x)=\frac{x^3}{3}+C_1\). Using \(f'(0)=7\), we get \(C_1=7\). Integrate again: \(\displaystyle f(x)=\frac{x^4}{12}+7x+C_2\). Using \(f(0)=2\), we get \(C_2=2\). Thus \(\displaystyle f(x)=\frac1{12}x^4+7x+2\). Answer: B.
