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Questions tagged [integration-by-parts]

To be used when the technique of Integration By Parts is the dominant topic of the question.

The integration by parts technique is used frequently. The method is used when integrating the product of functions by using an identity that is the result taking the integral of the multiplication rule for derivatives. The main goal of it is to change the integration so that one of the functions inside is integrated while the other is differentiated. Repeated application is intended to make one of the functions reduce to a constant, while having the other function be something that loops as it is repeatedly integrated, such as $\sin(x)$, $\cos(x)$, and $e^x$

The term LIATE is generally used to determine which item should be differentiated. Higher items should take precedence:

  • L - logarithmic functions
  • I - inverse trigonometric functions
  • A - algebraic functions
  • T - trigonometric functions
  • E - exponential functions

An outline of the proof for integration by parts is given as follows:

Take the multiplication rule for derivatives:

$$\frac {\mathrm{d}}{\mathrm{d}x}f(x)g(x) = f'(x)g(x) + f(x)g'(x)$$

Shift terms around:

$$f'(x)g(x) = \frac {\mathrm{d}}{\mathrm{d}x}f(x)g(x) - f(x)g'(x)$$

Integrate both sides:

$$\int f'(x)g(x) dx = f(x)g(x) - \int f(x)g'(x) dx$$

And that final line is the identity known as integration by parts.

229 questions
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Integration by parts, the cases when it does not matter what $u$ and $dv$ we choose.

I was reviewing Integration By Parts on Brilliant.org where an example they use is $$\int x \ln x \;dx$$ Let $u=\ln x$ and $dv=x\;dx$ such that $$\begin{align} \int x \ln x\;dx&\;=\;\frac 12x^2\ln x\;-\;\frac 12\int \frac{x^2}x\;dx\\ \\ &\;=\;…
user408202
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Methods for choosing $u$ and $dv$ when integrating by parts?

When doing integration by parts, how do you know which part should be $u$ ? For example, For the following: $$\int x^2e^xdx$$ $u = x^2$? However for: $$\int \sqrt{x}\ln xdx$$ $u = \ln x$? Is there a rule for which part should be $u$ ? As this is…
Dan
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$\int (x^2+1)/(x^4+1)\ dx$

I have Divided the numerator and Denominator by $x^2$ to get $\dfrac{1+x^{-2}}{x^2+x^{-2}}$ then changed it into $(1+(x^{-2}))/[(x-x^{-1})^2 +2]$ then took $x-(1/x)$ as $u$ and Differentiated it with respect to $x$ to get $dx=du/(1+x^{-2})$ Finally…
user126358
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How to explain why Integration by parts apparently "fails" in the case of $\int \frac{f'(x)}{f(x)}dx$ without resorting to definite integrals?

Integrating by parts the following integral $$I=\int \frac{f'(x)}{f(x)}dx$$ gives us $$\begin{align*} I&=\int \frac{f'(x)}{f(x)}\,dx\\ &=\int\frac1{f(x)}f'(x)\,dx\\ &=\frac1{f(x)}f(x)-\int\left(\frac1{f(x)}\right)'f(x)\,dx\\ &=1+\int…
MathematicianByMistake
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How to compute $\int_0^{\frac{\pi}{2}} \frac{\ln(\sin(x))}{\cot(x)}\ \text{d}x$

I am trying to compute this integral. $$\int_0^{\frac{\pi}{2}} \frac{\ln(\sin(x))}{\cot(x)}\ \text{d}x$$ Any thoughts will help. Thanks.
mnm
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Use integration by parts to find the integral $\int\frac{\sqrt {4x^2-9}}{x^2}dx$

$$\int\frac{\sqrt {4x^2-9}}{x^2}dx$$ I tried to solve this using integration by parts, but I come up with something that is much more difficult to solve. How can this be solved?
Erik Hambardzumyan
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Per partes integration: $\int x^2 \ln x dx$

Can I asked you guys for help with solving of this integral? Thank you. It could go well with the per-parted method, but I am not able to finish it. $$\int x^2 \ln x dx$$
user2932090
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$\int \sec^3x dx$ in disguise

I found this integral, $$\int \sqrt{x^2+1}dx$$ on a problem list and I think it is a sneaky way of hiding a $\int \sec^3xdx$ problem but I am not sure if what I did was correct though, because of what happens at the end. So what I did was use…
user409521
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For continuous 1-periodic f. $\int_1^{\infty}\frac{f(x)}{x}dx $ converges.

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ is continuous and 1 - periodic. If $ \int_0^1f(x)dx = 0$. Then $$\int_1^{\infty}\frac{f(x)}{x}dx $$ converges. My question is that does a bound on the $\int_1^{\infty}\frac{f(x)}{x}dx$ will imply convergence…
na1201
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For large $n$, show that $\int\limits_{0}^{1}\frac{nx^{n-1}dx}{1+x^2} $ nearly equals $\frac{1}{2}$.

For large $n$, show that $$\int\limits_{0}^{1}\frac{nx^{n-1}dx}{1+x^2} $$ nearly equals $\frac{1}{2}$. Integrating by parts we get $$\int\limits_{0}^{1}\frac{nx^{n-1}dx}{1+x^2}=\Bigg(\frac{x^n}{1+x^2}\Bigg)^{1}_{0} - \int\limits_{0}^1…
mathemather
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Solving integration by parts $\int\frac{xe^{2x}}{(1+2x)^2}\,dx$

$\displaystyle\int\frac{xe^{2x}}{(1+2x)^2}\,dx$ With this particular problem. my approach is to to rewrite the integral as $$\int xe^{2x}\frac{1}{(1+2x)^2}\,dx$$ and then pick a $u$ and a $dv$ and take it from there. The only issue I'm running into…
wa7d
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How do I integrate $\int\cos^2(x)\sin^\frac{2-\alpha}{\alpha-1}(x)\ dx$

I am having trouble solving this integral. $$\int\cos^2(x)\sin^\frac{2-\alpha}{\alpha-1}(x)\ dx$$ This is what I have done so far using integration by parts: Let $u=\sin^\frac{2-\alpha}{\alpha-1}(x)$ and $v'=\cos^2(x)$. Then we…
Joe Taylor
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compute $\int_0^1 {{x^{k - 1}}{e^{ - x}}} dx$

compute $\int_0^1 {{x^{k - 1}}{e^{ - x}}} dx$ This what I did: By integration by part : $\int_0^1 {{x^{k - 1}}{e^{ - x}}} dx = \frac{{{e^{ - 1}}}}{k} + \frac{1}{k}\int_0^1 {{x^k}{e^{ - x}}} $ I'am stuck here Some help would be appreciated
bill kid
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Deriving Yang-Mills equations

On the same spirit of this unanswered question I am proposing this question which I have been trying for some time now. Here I'm working with dimension $n = 4$ (identifying $\mathbb H = \mathbb R^4$) considering the principal $SU(2)$-bundle with…
Aaron Maroja
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Integration by parts: "math is broken"

just trying to solve a small example on integration by parts, and a weird thing happens: I come to an original expression increased by one. Please help me find out where the flaw is! The task is to calculate the following indefinite…
weekens
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