✅ Manual Solution:

Let \(u = \frac{1}{x} = x^{-1}\)

Differentiating \(u\) with respect to \(x\) : \( du = -\frac{1}{x^2} dx\)

Substitute \(u\) and \(du\) into the integral:

Using the standard integral formula \(\displaystyle \int a^u du = \frac{a^u}{\ln a} + C\) (with \(a=2\)):

Hence

Determining the Value of \(k\)

By comparing our result with the given form \(\displaystyle k \cdot 2^{\frac{1}{x}} + C\):

The value of \(k\) is:

Since \(\log 2\) in the options typically denotes the natural logarithm \(\ln 2\) in calculus, the answer is:

\(\displaystyle \frac{-1}{\log 2}\)

✅ Manual Solution:

$$\cos 2x - \cos 2\alpha = (2\cos^2 x - 1) - (2\cos^2 \alpha - 1)$$$$\cos 2x - \cos 2\alpha = 2\cos^2 x - 2\cos^2 \alpha$$$$\cos 2x - \cos 2\alpha = 2(\cos^2 x - \cos^2 \alpha)$$ $$=2(\cos x - \cos \alpha)(\cos x + \cos \alpha)$$ Hence $$\frac{\cos 2x-\cos 2\alpha}{\cos x-\cos \alpha} = \frac{2(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x-\cos \alpha}$$ $$ = 2(\cos x + \cos \alpha)$$ Hence integral becomes:$$I = \int 2(\cos x + \cos \alpha) dx$$ $$I = \mathbf{2(\sin x + x \cos \alpha) + C}$$

🤖 AI Generated Solution:

We use partial fraction decomposition for the given integrand. $$ \frac{x^2+x+1}{(x+2)(x^2+1)} = \frac{A}{x+2} + \frac{Bx+C}{x^2+1} $$ Multiplying both sides by $(x+2)(x^2+1)$, we get: $$ x^2+x+1 = A(x^2+1) + (Bx+C)(x+2) $$ Comparing coefficients of powers of $x$: For $x^2$: $A+B=1$ For $x$: $2B+C=1$ Constant: $A+2C=1$ Solving these equations gives $A=\frac{3}{5}$, $B=\frac{2}{5}$, $C=\frac{1}{5}$. Substitute these values back into the decomposition: $$ \int\left(\frac{3/5}{x+2} + \frac{(2/5)x+1/5}{x^2+1}\right)dx $$ $$ = \frac{3}{5}\int\frac{1}{x+2}dx + \frac{1}{5}\int\frac{2x}{x^2+1}dx + \frac{1}{5}\int\frac{1}{x^2+1}dx $$ Each integral can be evaluated directly: $$ = \frac{3}{5}\ln|x+2| + \frac{1}{5}\ln(x^2+1) + \frac{1}{5}\tan^{-1}(x) + C $$ The final answer is $\boxed{\frac{3}{5}\ln|x+2| + \frac{1}{5}\ln(x^2+1) + \frac{1}{5}\tan^{-1}(x) + C}$.

🤖 AI Generated Solution:

Let the given integral be $I$. $$I = \int\frac{\cos x}{(4+\sin^{2}x)(5-4\cos^{2}x)}dx$$ Let $t = \sin x$. Then $dt = \cos x \, dx$. Also, $\cos^{2}x = 1 - \sin^{2}x = 1 - t^{2}$. Substitute these into the integral: $$I = \int\frac{dt}{(4+t^{2})(5-4(1-t^{2}))}$$ $$I = \int\frac{dt}{(4+t^{2})(5-4+4t^{2})}$$ $$I = \int\frac{dt}{(4+t^{2})(1+4t^{2})}$$ Now, we use partial fraction decomposition for $\frac{1}{(4+t^{2})(1+4t^{2})}$. Let $\frac{1}{(4+t^{2})(1+4t^{2})} = \frac{A}{4+t^{2}} + \frac{B}{1+4t^{2}}$. Multiplying both sides by $(4+t^{2})(1+4t^{2})$, we get: $1 = A(1+4t^{2}) + B(4+t^{2})$ To find $A$ and $B$, we can set $t^2$ to specific values: Set $t^{2}=-4$: $1 = A(1+4(-4)) + B(0) \implies 1 = -15A \implies A = -\frac{1}{15}$. Set $t^{2}=-\frac{1}{4}$: $1 = A(0) + B(4-\frac{1}{4}) \implies 1 = B(\frac{15}{4}) \implies B = \frac{4}{15}$. So, the integral becomes: $$I = \int\left(\frac{-\frac{1}{15}}{4+t^{2}} + \frac{\frac{4}{15}}{1+4t^{2}}\right)dt$$ $$I = -\frac{1}{15}\int\frac{1}{2^{2}+t^{2}}dt + \frac{4}{15}\int\frac{1}{1+(2t)^{2}}dt$$ Using the standard integral formula $\int\frac{1}{a^{2}+x^{2}}dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C$: For the first term: $-\frac{1}{15} \cdot \frac{1}{2}\arctan\left(\frac{t}{2}\right) = -\frac{1}{30}\arctan\left(\frac{t}{2}\right)$. For the second term, let $u=2t$, so $du=2dt \implies dt=\frac{1}{2}du$: $\frac{4}{15}\int\frac{1}{1+u^{2}}\frac{1}{2}du = \frac{2}{15}\int\frac{1}{1+u^{2}}du = \frac{2}{15}\arctan(u) = \frac{2}{15}\arctan(2t)$. Combining both parts and substituting $t=\sin x$ back: $$I = -\frac{1}{30}\arctan\left(\frac{\sin x}{2}\right) + \frac{2}{15}\arctan(2\sin x) + C$$ The final answer is $\boxed{-\frac{1}{30}\arctan\left(\frac{\sin x}{2}\right) + \frac{2}{15}\arctan(2\sin x) + C}$.

🤖 AI Generated Solution:

To find the integral, we first decompose the rational function into partial fractions. 1. **Partial Fraction Decomposition**: Let $\frac{x^{2}+1}{(x-1)^{2}(x+3)} = \frac{A}{x-1} + \frac{B}{(x-1)^{2}} + \frac{C}{x+3}$. Multiplying by $(x-1)^{2}(x+3)$ gives: $x^{2}+1 = A(x-1)(x+3) + B(x+3) + C(x-1)^{2}$. 2. **Find the coefficients A, B, C**: * Set $x=1$: $1^2+1 = B(1+3) \implies 2 = 4B \implies B = \frac{1}{2}$. * Set $x=-3$: $(-3)^2+1 = C(-3-1)^2 \implies 10 = 16C \implies C = \frac{10}{16} = \frac{5}{8}$. * Set $x=0$: $0^2+1 = A(-1)(3) + B(3) + C(-1)^2 \implies 1 = -3A + 3B + C$. Substitute $B=\frac{1}{2}$ and $C=\frac{5}{8}$: $1 = -3A + 3\left(\frac{1}{2}\right) + \frac{5}{8}$ $1 = -3A + \frac{3}{2} + \frac{5}{8} \implies 1 = -3A + \frac{12+5}{8} \implies 1 = -3A + \frac{17}{8}$. $3A = \frac{17}{8} - 1 = \frac{9}{8} \implies A = \frac{3}{8}$. 3. **Integrate the partial fractions**: The integral becomes $\int \left(\frac{3/8}{x-1} + \frac{1/2}{(x-1)^{2}} + \frac{5/8}{x+3}\right)dx$. $\int\frac{x^{2}+1}{(x-1)^{2}(x+3)}dx = \frac{3}{8}\int\frac{1}{x-1}dx + \frac{1}{2}\int(x-1)^{-2}dx + \frac{5}{8}\int\frac{1}{x+3}dx$. 4. **Perform the integration**: $= \frac{3}{8}\ln|x-1| + \frac{1}{2}\left(\frac{(x-1)^{-1}}{-1}\right) + \frac{5}{8}\ln|x+3| + C$ $= \frac{3}{8}\ln|x-1| - \frac{1}{2(x-1)} + \frac{5}{8}\ln|x+3| + C$. The final answer is $\boxed{\frac{3}{8}\ln|x-1| - \frac{1}{2(x-1)} + \frac{5}{8}\ln|x+3| + C}$.