# Tag Info

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Logs and inverse trig functions, when part of a more complicated integral, tend to cause problems. By differentiating them, you get 'simpler' more algebraic expressions, because $$\frac{d}{dx}\ln x=\frac{1}{x}$$ and $$\frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^2}}$$ (the other inverse trig functions have similar derivatives with square roots in them). If you ...

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Put $$\frac{1}{x-1}=t$$ $\implies$ $$\frac{dx}{(x-1)^2}=-dt$$ So $$-I=\int \frac{t \,dt}{\sqrt{7t^2+8t+1}}$$ Put $$7t^2+8t+1=z^2$$ $\implies$ $$(7t+4)dt=zdz$$ So $$-7I=\int \frac{7t+4-4 \,dt}{\sqrt{7t^2+8t+1}}=\int \frac{7t+4 \,dt}{\sqrt{7t^2+8t+1}}-\int \frac{4 \,dt}{\sqrt{7t^2+8t+1}}$$ So $$-7I=z-\int \frac{4 ... 2 Let us consider two integrals$$I_1=\int \frac{\mathrm{d}x}{f(x)+1}I_2=\int \frac{\mathrm{d}y}{g(y)+c}$$If it is possible to make a change of variable such that$$g(y)+c=f(x)+1$$that is to say$$y=g^{-1}(f(x)+1-c)$$the denominators of the integrands will become the same. However, we shall have the problem of \frac {dy}{dx}; if this is a constant, ... 0 You're going in the wrong direction. Hint: Rewrite the integral as \frac{1}{9} \displaystyle \int \frac{1}{(1 + \frac{4}{9} x^2)} dx Let u = \frac{2}{3} x. 0$$\int\frac{dx}{9+4x^2}=\frac19\int\frac{dx}{1+\left(\frac{2x}3\right)^2}$$Use Trigonometric substitution,$$\frac{2x}3=\tan\theta$$0 Let u=2x+1. Then \text du=2$$\int \sec(2x+1) \ \text dx=\int \frac{\sec(u)}{2} \ \text du=\frac 12\int \sec(u) \, \text du=\frac 12\ln|\sec(u)+\tan(u)|+\text C$$Reverse the substitution$$\frac 12\ln|\sec(2x+1)+\tan(2x+1)|+\text C\color{green}{\int \sec(2x+1) \, \text dx =\frac 12\ln|\sec(2x+1)+\tan(2x+1)|+\text C}$$0$$\int\sec({2x+1})\,dx=\frac{1}{2}\int\sec({2x+1})\,d(2x+1)$$Can you take it from here? 1 Set 2x+1=u\implies 2dx=du$$\int\sec(2x+1)\ dx=\frac12\int\sec u\ du=\frac{\ln|\sec u+\tan u|}2+K$$Replace back u with 2x+1 0 You neglected the chain rule:$$ \frac{d}{dx} \ln|\sec(2x+1)+\tan(2x+1)| = \sec(2x+1)\cdot 2. $$-1$$\int\frac{x^2 + 1}{x^3 + 3x + 1} dx=\dfrac{1}{3}\int\frac{3(x^2 + 1)}{x^3 + 3x + 1} dx=\\ \dfrac{1}{3}\int\frac{1}{x^3 + 3x + 1} d(x^3 + 3x + 1)=\dfrac{1}{3}\ln\left|x^3 + 3x + 1\right|+C$$0 f=\Phi^{-1}(g(x)) for any function g with a closed form taking values between 0 and 1 will work. Or use integration by parts to obtain$$b\Phi\bigl(f(b)\bigr)-a\Phi\bigl(f(a)\bigr)-\int_a^b x f'(x) \phi\bigl(f(x)\bigr) dx,$$which suggests that your problem simplifies to finding values such that the sum of the first two terms has a closed form ... 0 This is the Parallel axis theorem, also known as Steiner's Rule. You can read up on it here: http://en.wikipedia.org/wiki/Parallel_axis_theorem 1 Let \displaystyle a=\frac{7-t^2}{t^2-1} instead of \displaystyle a=\frac{7-t}{t-1}  ?? Doing that and simplifying we get, \displaystyle -\frac{12 t}{\left(t^2-1\right)^2} and integrating, we get$$-2 \left(-\frac{3 t}{7 \left(t^2-7\right)}+\frac{2 \log \left(\sqrt{7}-t\right)}{7 \sqrt{7}}-\frac{2 \log \left(t+\sqrt{7}\right)}{7 \sqrt{7}}\right)$$... 1 You can use the identity that$$\int e^{kx}dx=\frac{1}{k}e^{kx}+C$$for complex constants k and C. Thus the last integral can be evaluated as follows:-$$\int_{-\infty}^\infty e^{-a|\gamma|} e^{i\gamma(x-vt)} d\gamma = \int_{-\infty}^0 e^{a\gamma} e^{i\gamma(x-vt)} d\gamma+\int_{0}^\infty e^{-a\gamma} e^{i\gamma(x-vt)} d\gamma\\=\int_{-\infty}^0 ...

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Yes, your work with integration by parts shows that for even powers $t^{2n}$, there is no elementary antiderivative for $t^{2n} e^{-ct^2}$ because there is no elementary antiderivative for $e^{-ct^2}$.

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What you wrote, integrating by parts, is kind of correct. The integral of the Gaussian is impossible to express in terms of simple functions, but is well known, it is called error function. It has the defining property: $$\mathrm{erf}\bigg(\frac{x}{\sqrt{2}}\bigg) =\sqrt{\frac{2}{\pi}} \int_0^x e^{-\frac{1}{2}t^2}\,dt.$$ Therefore the antiderivative ...

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There is no elementary anti-derivative for $e^{-t^2/2}$, but there are several special functions such as the error function $$\int e^{-t^2/2}\,\mathrm{d}t=\sqrt{\frac\pi2}\,\mathrm{erf}\left(\frac t{\sqrt2}\right)+C$$ Note that $\mathrm{erf}(0)=0$. Your integration by parts is correct. The error function above is defined as $$... 1 Using IBP by letting u=t, du=dt, and$$ dv=te^{-\large\frac12t^2}\ dt\quad\Rightarrow\quad v=\int te^{-\large\frac12t^2}\ dt=-e^{-\large\frac12t^2}, $$yield$$ \int t^2e^{-\large\frac12t^2}\ dt=-te^{-\large\frac12t^2}+\int e^{-\large\frac12t^2}\ dt=-te^{-\large\frac12t^2}+\sqrt{\frac{\pi}{2}}\text{erf}\left(\frac{x}{\sqrt{2}}\right). $$The last part is ... 0 Since I spent so much time trying to work this out, I'm going to type the answer even though there are already two perfectly good answers. Hah! Anyway, here's how to do it. Let$$u = b - a \cos x$$and let$$y^2 = a^2 - b^2 + 2ub = a^2 + b^2 - 2ab \cos x.$$Notice that then making the change of variables to y, we have that$$ \begin{align*} 2ydy = ...

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Another hint : \begin{align} \frac{3x^{2}-x+2}{x-1}&=\frac{3x^{2}-3x+2x+2}{x-1}\\ &=\frac{3x^{2}-3x}{x-1}+\frac{2x}{x-1}+\frac{2}{x-1}\\ &=\frac{3x(x-1)}{x-1}+2\left(\frac{x-1+1}{x-1}\right)+\frac{2}{x-1}\\ &=3x+2\left(1+\frac{1}{x-1}\right)+\frac{2}{x-1}\\ \end{align}

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