Questions tagged [indefinite-integrals]

Question about finding the primitives of a given function, whether or not elementary.

The indefinite integral is defined as a set of all functions $F$ such that $F' = f$. Each member of the set is called an antiderivative. For example, $$\int f(x) dx = \lbrace F(x): F'(x) = f(x) \rbrace$$ also commonly denoted as $$F(x) + C.$$

If $F'(z) = f(z)$ then we denote

$$\int f(z) \; dz = F(z)$$

and call $F(z)$ a primitive of $f(z)$, also called an antiderivative. This result, while taught early in elementary calculus courses, is actually a very deep result connecting the purely algebraic indefinite integral and the purely analytic (or geometric) definite integral.

Since the derivative of a constant is zero, any constant may be added to an antiderivative and will still correspond to the same integral. Another way of stating this is that the antiderivative is a nonunique inverse of the derivative. For this reason, indefinite integrals are often written in the form $$\int f(z)\;dz=F(z)+C$$

where $C$ is an arbitrary constant known as the constant of integration.

It may happen that there is no elementary function$^1$ such that $$\int f(z) \; dz = F(z)$$ In such case, we define a new function which is not elementary but still satisfies our definition. For example, there is no elementary function $F$ such that $F'(z) = \displaystyle \frac{e^z}{z}$. However, if we define

$$\int \frac{e^z}{z} dz = C + \log z + \int_0^z \frac{e^t-1}{t} dt$$

we can readily check that $F' = f$.

$^1$: A function built up of a finite combination of constant functions, field operations (addition, multiplication, division, and root extractions - the elementary operations) and algebraic, exponential, and logarithmic functions and their inverses under repeated compositions. See also.

Source: Wolfram Mathworld

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Indefinite integration of $\sqrt{f(x)/x^2}$

If $$f(x) =\frac {x+2}{2x+3}$$ then the integral $$ \int{\sqrt{\frac{f(x)}{x^2}}}\,dx$$ is equal to? The expression is way beyond my comprehension, no idea in this case.
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Hint to solve $\int \frac{1-x^2}{x^2+x-2}$

I need to solve this: $$\int \frac{1-x^2}{x^2+x-2}$$ I tried to simplify the Denominator in this way: $$ x^2+x-2 = (x-1)(x+2)$$ so my new integral is: $$\int \frac{1-x^2}{(x-1)(x+2)}$$ So now I have N°(x) > D°(x) and as far as I know in this case I…
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how i should solve this problem?

$\int \frac{\text{d}x}{(1+\sqrt{x})(x-x^2)}$ how i should solve this problem ? i think we should take $x=\sin^2(x)$ and then proceed but still not able to solve , please help !
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Solve $\int \frac{2x+1}{x^2-2x+2}$

I need to solve $$\int \frac{2x+1}{x^2-2x+2}dx$$ I have noticed that the Numerator is almost the derivate of the Denominator so I did this: $$\int \frac{2x-2+2+1}{x^2-2x+2}dx = \int \frac{2x-2}{x^2-2x+2}dx + \int \frac{3}{x^2-2x+2}dx$$ so the first…
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Solve this integral $\int \frac{x-1}{x+4x^3} \, dx$

I need to solve this integral $$\int \frac{x-1}{x+4x^3} \, dx$$ As some users suggest me in my previous question here I used the partial fractions. and I got: $$\int \frac{x-1}{x+4x^3} \, dx = - \int \frac{1}{x} \, dx + \int \frac{4x+1}{1+4x^2} \,…
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Is the expression $\int {1\over 2}E^2 \, dV$ infinite for infinite volume?

There exist a calculation about electromagnetic mass: $$m_\mathrm{em} = \int {1\over 2}E^2 \, dV = \int\limits_{r_e}^\infty \frac{1}{2} \left( {q\over 4\pi r^2} \right)^2 4\pi r^2 \, dr = {q^2 \over 8\pi r_e}$$ Reducing $r_e$ we get infinite mass.…
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Find the indefinite integral

$$\int \frac{x^3 - 2x +1}{\sqrt{x}} dx$$ First term: $x^3 = \frac{1}{4}x^4$ Second term: $2x = x^2$ Third term: $1 = x$ Fourth term: $\sqrt{x} = x^{1/2}$ I know the fourth term is wrong and forget how to solve that one but which other ones are…
alex
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How did they use the chain rule to seperate the variables ?

"You have 1/x dx/dy = 1/y, and you're assuming that x is a function of y. From this you can conclude that an antiderivative of the left hand side of the above differs from an antiderivative of the righthand side by a constant. This is saying: ∫1/x…
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How does this integral change due to restricted x values?

So, I integrated this integral: $$I = \int\frac{dx}{(x^2+4)\sqrt{1-x^2}}$$ And got: $$I = \frac{1}{2\sqrt{5}}\arctan{\frac{x\sqrt{5}}{2\sqrt{1-x^2}}}+C$$ But, since $-1 < x < 1$, how does it affect the integral?
A6SE
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Integral of $\sin(ax+b)\cos(ax+b)$

$\displaystyle\int \sin(ax+b)\cos(ax+b) \, dx$ is giving me two answers; $\dfrac{1}{2a}\sin^2(ax+b)+C$ negative of $\dfrac{1}{4a}\cos2(ax+b) +C$ Are these two results the same? Can you please prove it? Thank you
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Help me solve the indefinite integral

So, I ran across this in my workbook and have no idea where to start. I can't recognize any basic form or any substitution which I can make. $$I=\int\frac{dx}{\sqrt[n]{(x-a)^{n+1}(x-b)^{n-1}}}$$
A6SE
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Confusing answer for $\int\frac{x^4+1}{x^4-1}dx$

While prepairing for upcoming test I met this integral and was confused with the answer they got in the answer book for it. Please show me where I am wrong. $$\int\frac{x^4+1}{x^4-1}dx=\int\frac{x^4}{x^4-1}dx +…
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Indefinite Integral

I'm having problems with starting this integral, I'm not looking for the solution, but the first step/two would be appreciated. Integral: Evaluate $$ \int dx \sqrt{x(4-x)} $$
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Integrate $\int x^2e^{-x^2}dx$

I need your advice in integrating $\int x^2e^{-x^2}dx$ by parts. I went this way, $$\int x^2e^{-x^2}dx=\int x*xe^{-x^2}dx=\frac{1}{2}\int x e^{-x^2}dx^2$$ I think I need to substitute $x^2=t$. Is that correct?
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Advice for $\int\frac{\sin{2x}}{\sqrt{\cos^4{x}+3}}dx$

could anybody guide me to the next step $$\int\frac{\sin{2x}}{\sqrt{\cos^4{x}+3}}dx=\int\frac{2\sin{x}\cos{x}}{\sqrt{\cos^4{x}+3}}dx=-\int\frac{\cos{x}}{\sqrt{\cos^4{x}+3}}d{\cos^2{x}}=...$$ and here I stuck. I don't know how to get rid of $cos{x}$…