Linked Questions

First approach. $\int \frac{1}{1+x^2} dx=\frac{x}{1+x^2}+2\int \frac{x^2}{\left(1+x^2\right)^2} dx=\frac{x}{1+x^2}+2\int \frac{1}{1+x^2}dx-2\int \frac{1}{\left(1+x^2\right)^2}dx$ From this ...

Problem: Evaluate the following integral: \begin{eqnarray*} \int \frac{1}{(x^2+1)^2} \, dx \\ \end{eqnarray*} Answer: To do this, I let $x = \tan u$. Now we have $dx = \sec^2 u du$. \begin{eqnarray*} \...

I know that the integral looks like like the anti-derivative of $\arctan$, but i don't know how to use this fact so I tried to use a fraction decomposition to: $$\frac{1}{(x^2+1)^2}$$ to solve the ...

If any integration is in form $$\int \frac{1}{1+x^2}dx$$ it easily follows by substituting for $\tan^{-1}x$. But how to simplify if we have $$\int \frac{1}{(1+x^2)^2}dx$$

Suppose we know $\int \frac{1-x^2}{(x^2+1)^2}=\frac{x}{x^2+1}+C$ How to compute $\int \frac{1}{(x^2+1)^2}dx$? I tried writing it as $\frac{1+x^2-x^2}{(x^2+1)^2}=\frac{1-x^2}{(x^2+1)^2}+\frac{x^2}{(x^...

I worked differential linear equation and at end of equation I got this integral.Can someone give me a hint to do this: $$\int \frac{1}{\left(u^2+1\right)^2}du$$

How to solve for: $\int \frac{dx}{\left(x^2+4\right)^2}$ What I have tried so far: Let $x = 2 \tan \theta$ so that $dx = 2 \sec^2\theta d\theta$ $I = \int \frac{2 \sec^2 \theta d\theta}{\left(16 ...

Is there any possible way to calculate the integral of $\frac{1}{(x^2+1)^2}$ by partial fraction decomposition? I do not know the formulas for the trigonometric method. Thank you!

I've been tutoring some basic calculus, and it made me think about something pretty basic. Let me explain the problem by example: Say we are given the integral $\int \frac{x^2}{\sqrt{1-x^2}}\ \...

Express recurrence relation of the integral $$ I_n=\int\frac{dx}{(1+x^2)^n} $$ [My Answer] $$ I_n = \int\frac{1+x^2}{(1+x^2)^n}dx-\int\frac{x^2}{(1+x^2)^n}dx $$ $$ I_n=I_{n-1}-\int x\cdot\frac{x}{...

I am trying to find partial fractions of $\frac {1}{(x^2+1)^2}$. All the coefficients I get are zeros except the coefficient for the constant term which is 1, leaving me with the fraction I started ...

For my engineering math course I got a couple of exercises about indefinite integrals. I ran trought all of them but stumbled upon the following problem. $$\int \frac{1-x}{\sqrt{1+x-2x^2}}\,dx $$ ...

I am studying for a test and I am trying to evauate the integral below. I know how to simplify it with partial fractions, but when I try to solve it, I cannot seem to find a substitution that will ...

What is the easiest way to integrate $y=\frac{x+4}{\sqrt{-x^2-2x+3}}$ ? I tried to integrate it by making numerator in form: $-2x-2$ and then pulling it under differential, but the result drastically ...

$$\int \frac{1}{(x^2+1)^2}\mathrm dx$$ Look simple, but I stuck a little bit. What method is better in this situation? Please, help.