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  1. No $u$-substitution required for this integral. All you have to do is break the fraction down. This integral is actually equal to $\displaystyle\int\frac{x^{\frac{3}{2}}}{x^{\frac{1}{3}}}dx=\displaystyle\int\frac{x^{\frac{9}{6}}}{x^{\frac{2}{6}}}dx=\displaystyle\int x^{\frac{7}{6}}dx$. By utilizing the power rule, we can see that this integral evaluates to $\frac{6}{13}x^{\frac{13}{6}}+C$.

  2. No $u$-sub required here. We can just simplify this integral into a more familiar form by using trig identities: $\displaystyle\int\cot^2\theta d\theta=\displaystyle\int\csc^2\theta-1d\theta$. Now, for convenience, we can break this up into two separate integrals: $\displaystyle\int\csc^2\theta d\theta-\displaystyle\int1d\theta$. Now integrating, we get our answer of $-\cot\theta-\theta+C$, since the derivative of $\cot(x)$ is $-\csc^2(x)$.

  3. We can see that this integral is equal to $\displaystyle\int\frac{1}{\sin x+1}dx$. We can multiply by one in order to get the denominator into a trig identity: $\displaystyle\int\frac{1}{\sin x+1}\cdot\frac{1-\sin x}{1-\sin x}dx=\displaystyle\int\frac{1-\sin x}{\cos^2x}dx$. Separating the integrand, we get $\displaystyle\int\frac{1}{\cos^2x}dx-\displaystyle\int\frac{\sin x}{\cos^2x}dx$. Now solving, we get: $\displaystyle\int\sec^2xdx-\displaystyle\int\sec x\tan xdx=\tan x-\sec x+C$.

  4. The hard part about this integral is getting it into a form that is easy to deal with. This can be done by using some tricky algebra and trigonometric identities. By rearranging the integrand, we can see that this integral becomes: $\displaystyle\int\sin^2x\tan^x\sin x dx=\displaystyle\int(1-\cos^2x)(\sec^2x-1)\sin xdx$ (using trig identities). Now, by foiling the integrand, we get a sum of integrals that we can split into three separate integrals $\displaystyle\int\sec x\tan xdx+\displaystyle\int\cos^2x\sin xdx-2\displaystyle\int\sin xdx$. The two outer integrals seem easy enough, but in order to evaluate the one in the middle ($\displaystyle\int\cos^2x\sin xdx$) we need to use $u$-substitution. Set $u=\cos x$. Thus, $du=-\sin x$. Hence, the middle integral becomes $-\displaystyle\int u^2du$. Evaluating this gives $-\frac{u^3}{3}$ which, in terms of $x$, is $-\frac{\cos^3x}{3}$. Integrating the other two as well, we get the full answer: $\sec x+2\cos x-\frac{\cos^3x}{3}$.

  5. By using $u$-sub, we can set $u=x^3$ and $du=3x^2$. Thus we can rewrite the integral as $\frac{1}{3}\displaystyle\int e^udu$ (we multiply by $\frac{1}{3}$ in order to cancel out the coefficient of 3 in front of $du$). Now, integrating we get an answer of $\frac{e^u}{3}+C$, which, in terms of $x$, is actually $\frac{e^{x^3}}{3}+C$.

  6. If we choose $u=x^2+4$, then $du=2x$. Hence, the integral becomes $\displaystyle\int u^3du$. Here, since the power is a constant, we can implement the power rule for integrals, so this antiderivatives evaluates to $\displaystyle\frac{u^4}{4}+C$. Substituting back to get the answer in terms of $x$, we get $\displaystyle\int2x(x^2+4)^3dx=\frac{(x^2+4)^4}{4}+C$.

  7. For this problem, choose $u=\frac{x}{3} \to du=\frac{1}{3}$. We can rewrite this integral in terms of $u$, but we have to multiply by 3 in order to counterbalance the effect that $d=\frac{1}{3}$ has. So, the integral becomes $3\displaystyle\int\sec^2udu$. Using our previous knowledge of derivatives, we can see that this integral evaluates to $\frac{1}{3}\tan u+C$, which in terms of $x$ is equal to $\frac{1}{3}\tan\frac{x}{3}+C$.

  8. Choosing $u=x+2$ and $du=1$, we can also arrive at the conclusion that $x=u-2$. Now, this integral can be written in terms of $u$: $\displaystyle\int\frac{(u-2)^2+4}{u}du$. If we expand the numerator, we get $\displaystyle\int\frac{u^2-4u+8}{u}du$. Now separating the integrals, we get $\displaystyle\int\frac{u^2}{u}du-\displaystyle\int\frac{4u}{u}du+\displaystyle\int\frac{8}{u}du=$ $\displaystyle\int u du-\displaystyle\int4 du +8\displaystyle\int\frac{1}{u}du$. Now integrating all separately, we get $\frac{u^2}{2}-4u+8\ln|u|+C$, which is equal to $\frac{(x+2)^2}{2}-4(x+2)+8\ln|x+2|+C$.

  9. Choosing $u=x+1$, $du=1$, we can see that $x=u-1$. Thus, this integral becomes $\displaystyle\int u^3(u-1+3)du=\displaystyle\int u^4+2u^3du$. Breaking this apart, we get $\displaystyle\int u^4du+2\displaystyle\int u^3du$, which evaluates to $\frac{u^5}{5}+2\frac{u^4}{4+C}$. Substituting back, we get our answer to this integral: $\frac{(x+1)^5}{5}+\frac{(x+1)^4}{2}$.

  10. Using back substitution, we can set $u=x+3, du=1, x=u-3$. Thus, we can rewrite this integral in terms of $u$ as follows:$\displaystyle\int\frac{1}{2}\frac{u-3+5}{u}du=\frac{1}{2}\displaystyle\int\frac{u+2}{u}du$. Breaking this into two integrals, we get $\frac{1}{2}\displaystyle\int\frac{u}{u}du+\frac{2}{2}\displaystyle\int\frac{1}{u}du$ which is equal to $\frac{1}{2}\displaystyle\int 1du+\displaystyle\int\frac{1}{u}du$. Evaluating this, we get a result of $\frac{u}{2}+\ln|u|+C$. Thus, if we substitute back, we have our answer: $\frac{x+3}{2}+\ln|x+3|+C$.

  11. As is usually the case with rational functions with a variable denominator, we choose $u=\ln(x)$ and $du=\frac{1}{x}$. Thus, we can rewrite this integral in terms of $u$ as follows: $\displaystyle\int\frac{1}{4}(3+u)^2(2-u)du=\frac{1}{4}\displaystyle\int -u^3+4u^2+3u+18du$ (after the foiling is done). Splitting this into four separate integrals, we get $-\frac{1}{4}\displaystyle\int u^3du+\frac{1}{4}\displaystyle\int u^2du+\frac{3}{4}\displaystyle\int udu+\frac{1}{4}\displaystyle\int18du$ (notice how we pull out the coefficients from some expressions and multiply it with the fraction in front of the integral). Now, we can integrate. This evaluates to $-\frac{1}{4}\cdot\frac{u^4}{4}-\frac{u^3}{3}+\frac{3}{4}\cdot\frac{u^2}{2}+\frac{1}{4}\cdot18u+C$. Now, getting this answer in terms of $x$ by substituting back, we get our final answer: $-\frac{\ln^4x}{16}-\frac{\ln^3x}{3}+\frac{3\ln^2x}{2}+\frac{9\ln x}{2}+C$.

  12. If we choose $u=\ln x$ and $du=\frac{1}{x}$, then this integral becomes very easy: $\displaystyle\int\sin u du=-\cos u+C$. Subbing back in, we get the following for a final answer: $-\cos(\ln(x))+C$.

  13. Substituting $u=\ln(x)$ and $du=\frac{1}{x}$, we can rewrite this integral as $\displaystyle\int\frac{3}{u}du=3\displaystyle\int\frac{1}{u}du$. Integrating, we get $3\ln|u|+C$; subbing back in, we get $3\ln|\ln(x)|+C$.

  14. Immediately, we can separate this into two different integrals: $\displaystyle\int e^{-x}dx+\displaystyle\int\sec^2\left(\displaystyle\frac{x}{10}\right)dx$. Since these are now two separate integrals, we can choose two separate $u$ values for each of them. For $\displaystyle\int e^{-x}dx$, choose $u=-x, du=-1$; for $\displaystyle\int\sec^2\left(\displaystyle\frac{x}{10}\right)dx$, choose $u=\displaystyle\frac{x}{10}, du=\displaystyle\frac{1}{10}$. Now, we can continue with the substitution: $-\displaystyle\int e^u du+10\displaystyle\int\sec^2(u)du$. Evaluating, we get $-e^u+10\tan u+C$. Now, the trick is to substitute back the correct things for each equation. If we do this, we get the answer: $-e^{-x}+10\tan\displaystyle\frac{x}{10}+C$.