Author: Study Clue

Evaluate ∫01×31+x2 dx∫011+x2x3dx using a smart algebraic split before integrating. A 12+12ln⁡221+21ln2 B 12−12ln⁡221−21ln2 C ln⁡2−12ln2−21 D 14ln⁡241ln2 Explanation Write x31+x2=x−x1+x21+x2x3=x−1+x2x.

Evaluate ∫4x(1+2×2) dx∫(1+2×2)4xdx using a suitable substitution. A ln⁡(1+2×2)+Cln(1+2×2)+C B ln⁡∣x∣+Cln∣x∣+C C 12ln⁡(1+2×2)+C21ln(1+2×2)+C D 14ln⁡(1+2×2)+C41ln(1+2×2)+C Explanation Let u=1+2x2u=1+2×2. Then du=4x dxdu=4xdx. So

A function f(x)f(x) satisfies f′(x)=2x(1+x2)4f′(x)=2x(1+x2)4. What is ∫2x(1+x2)4 dx∫2x(1+x2)4dx? A (1+x2)55+C5(1+x2)5+C B (1+x2)5+C(1+x2)5+C C 2x(1+x2)4+C2x(1+x2)4+C D (1+x2)44+C4(1+x2)4+C Explanation Use substitution u=1+x2u=1+x2,

When you see ∫f′(x) f(x) dx∫f′(x)f(x)dx, which method is usually simplest to start with? A By parts method B Partial fractions C

Which statement best defines an antiderivative of f(x)f(x)? A Function is constant B Integral equals zero C Derivative equals f(x)f(x)

Check whether lim⁡z→0zˉzlimz→0zzˉ exists in complex sense A Equals 11 B Equals −1−1 C Does not exist D Equals 00

Evaluate lim⁡z→0ez−1zlimz→0zez−1 using series idea A 00 B ee C 11 D Does not exist Explanation Using ez=1+z+z22!+⋯ez=1+z+2!z2+⋯, we get

For f(z)=z2−1z−1f(z)=z−1z2−1, what is lim⁡z→1f(z)limz→1f(z)** after simplification** A 11 B 00 C 22 D Does not exist Explanation Factor z2−1=(z−1)(z+1)z2−1=(z−1)(z+1).

While checking lim⁡z→af(z)limz→af(z), which approach best confirms path independence A Compare two different paths B Check only real axis C

Which condition must hold for lim⁡z→af(z)limz→af(z) to exist in the complex plane A Same on real axis B Same on