Articles of indefinite integrals

Evaluate $\int \dfrac{1}{\sqrt{1-x}}\,dx$

Find $$\int \dfrac{1}{\sqrt{1-x}}\,dx$$ I did this and got $\dfrac23(1-x)^{\frac32} + c$ But a online calculator is telling me it should be $-2(1-x)^{\frac12}$ What one is on the money and if not me why?

Integrating $\frac{1}{(x^4 -1)^2}$

How to solve the the following integral? $$\int{\frac{1}{(x^4 -1)^2}}\, dx$$

Can indefinite double integrals be solved by change of variables technique?

I have a simple double integral: $$\int\int(x+y)dxdy$$ Now, change the variables according to the following scheme: $$x=u+v$$ $$y=u-v$$ The jacobian is then: $$|\frac{\partial (x,y)}{\partial (u,v)}|=1$$ Substitute u and v into the integral, solving the integral and substituting y and x back in their place equals: $$(\frac{x-y}{2})(\frac{x+y}{2})^2$$ Which is obviously not the same as the result […]

Integrate:$\int({1+x-\frac{1}{x}})e^{x+\frac{1}{x}}dx$

How to find $$\int({1+x-\frac{1}{x}})e^{x+\frac{1}{x}}dx$$ I am thinking about breaking it into the form $$\int{e^x (f(x)+f'(x))}dx=e^x f(x)+C$$ Don’t know how to split it. Please help.

Finding the antiderivative of the product of two functions given only their derivative properties

let $\alpha'(x)=\beta(x), \beta'(x)=\alpha(x)$ and assume that $\alpha^2 – \beta^2 = 1$. how would I go about calculating the following anti derivative : $\int (\alpha (x))^5 (\beta(x))^4$d$x$. Thank you.

Evaluating $ \int \frac{1}{5 + 3 \sin(x)} ~ \mathrm{d}{x} $.

What is the integral of: $\int \frac{1}{5+3\sin x}dx$ My attempt: Using: $\tan \frac x 2=t$, $\sin x = \frac {2t}{1+t^2}$, $dx=\frac {2dt}{1+t^2}$ we have: $\int \frac{1}{5+3\sin x}dx= 2\int \frac 1 {5t^2+6t+5}dt $ I’ll expand the denominator: $5t^2+6t+5=5((t+\frac 3 5 )^2+1-\frac 1 4 \cdot (\frac 6 5)^2)=5((t+\frac 3 5)^2+0.64)$. So: $2\int \frac 1 {5t^2+6t+5}dt = \frac […]

Integrating $\int\frac{xe^{2x}}{(1+2x)^2} dx$

I need help in integrating $$\int \frac{x e^{2x}}{(1+2x)^2} dx$$ I used integration by parts to where $u=xe^{2x}$ and $dv=\frac{1}{(1+2x)^2}$ to obtain $$=\frac{-1}{2(1+2x)}\left[e^{2x}+2xe^{2x}\right] + \frac{1}{2}\int \frac{1}{(1+2x)}\left(2e^{2x}+\frac{d}{dx}\left(2xe^{2x}\right)\right)dx$$ Which is pretty long, which made me question whether I’m doing it correctly or not. Are there any other ways to properly evaluate the integral? Do note though that I […]

Computing $\int (1 – \frac{3}{x^4})\exp(-\frac{x^{2}}{2}) dx$

How does one compute $$\int \left(1 – \frac{3}{x^4}\right)\exp\left(-\frac{x^{2}}{2}\right) dx?$$ Mathematica gives $(x^{-3} – x^{-1})\exp(-x^2/2)$ which indeed the correct answer, but how does one get there? Integration by parts gives $(y + y^{-3})e^{-y^2/2} + \int (y^2 + y^{-2})e^{-y^2/2} dy$, but I’m not sure what to do with the integral.

Closed form for $\int \frac{1}{x^7 -1} dx$?

I want to calculate: $$\int \frac{1}{x^7 -1} dx$$ Since $\displaystyle \frac{1}{x^7 -1} = – \sum_{i=0}^\infty x^{7i} $, we have $\displaystyle(-x)\sum_{i=0}^\infty \frac{x^{7i}}{7i +1} $. Is there another solution? That is, can this integral be written in terms of elementary functions?

Integration by Parts? – Variable Manipulation

$$\int x^3f”(x^2)\,\mathrm{d}x$$ Solve using Integration by Parts. \begin{align} u&=x^3\qquad\mathrm{d}v=f”(x^2) \\ \mathrm{d}u&=3x^2\qquad v=f'(x^2) \\ &=x^3f'(x)-\int f'(x^2)3x^2 \\ u&=3x^2\quad\mathrm{d}v=f'(x^2) \\ \mathrm{d}u&=6x \qquad v=f(x^2) \\ &=x^3f'(x^2)-[3x^2f(x^2)-\int f(x^2)6x] \end{align} No clue what do from here as the correct answer is: $$\frac{1}{2}(x^2f'(x^2)-f(x^2))+C$$ Can you guys think of anything? I appreciate the help. 🙂