Free integral calculator - solve indefinite, definite and multiple integrals with all the steps. Type in any integral to get the solution, steps and graph.

Enter the function you want to integrate into the Integral Calculator. Skip the f (x)= part and the differential dx! The Integral Calculator will show you a graphical version of your input while you type. Make sure that it shows exactly what you want. Use parentheses, if necessary, e.g. a/ (b+c).

This calculator computes the definite and indefinite integrals (antiderivative) of a given function with respect to a variable x. Input function: ? supported functions: sqrt, ln , e, sin, cos, tan, asin, acos, atan,... Solve: Find more worked-out examples in our database of solved problems.

Enter the function you want to integrate into the editor. The Integral Calculator solves an indefinite integral of a function. You can also get a better visual and understanding of the function and area under the curve using our graphing tool. Integration by parts formula: ?udv = uv−?vdu? u d v = u v -? v d u. Click the blue arrow to submit.

• ln^2 (x), ln (x)^2 — \ (\ln^2\left (x\right)\) Step-by-step integral calculator. With convenient input and explanations!

2025年3月11日 · C. 思路：. 贪心构造出一个数让整个序列满足要求，每个数之间都不相同，并且满足 {a_1=a2-a_3+a_4-a_5}+...+a_{2n}-a_{2n+1},我们可以将给定的数组中的最大的数成为 {a_1} ，删除的数可以用那个式子推出来，因为每个数都不能相同，我们可以使前面的n-1对全部构成负数，让最后的一对成正数，这样我们构造的 ...

Free Integral Calculator helps you solve definite and indefinite integration problems. Also double, triple and improper integrals. Answers, graphs, alternate forms.

We can use the definite integral: To evaluate this, we first find the antiderivative of x^2 x2, which is \frac {x^ {2+1}} {2+1}=\frac {x^3} {3} 2+1x2+1 = 3x3. Now, we apply the Fundamental Theorem of Calculus: So the area under the curve f (x)=x^2 f (x) = x2 between x=1 x = 1 and x=3 x = 3 is 26 26 square units.

2013年5月15日 · Do you know a trigonometric function that you might multiply by a a and substitute for x x? The idea is to get a square under the square root. I will give you a proof of how they can get the formula above.

x^2: x^{\msquare} \log_{\msquare} \sqrt{\square} \nthroot[\msquare]{\square} \le \ge \frac{\msquare}{\msquare} \cdot \div: x^{\circ} \pi \left(\square\right)^{'} \frac{d}{dx} \frac{\partial}{\partial x} \int \int_{\msquare}^{\msquare} \lim \sum \infty \theta (f\:\circ\:g) f(x)