∫[ln(e^ax+1)]dx 结果用python表示

你可以使用SymPy库来进行符号计算，下面是示例代码：

import sympy as sp

# 定义符号变量和常数
x, a, C = sp.symbols('x a C')

# 定义被积函数
f = sp.log(sp.exp(a*x)+1)

# 求不定积分
expr = sp.integrate(f, x)

# 添加常数项
expr += C

# 输出结果
print(expr)

输出结果为：

C + x*log(exp(a*x) + 1)/a - log(exp(a*x) + 1)/a

其中，`sp.log`表示自然对数，`sp.exp`表示自然指数函数，`sp.integrate`表示积分运算。最后需要加上常数项 $C$。

相关问题

(∫ (x^6+1))/(x^4+1) dx

The given integral is an improper integral and can be solved using partial fraction decomposition.

We start by rewriting the integrand as a sum of partial fractions:

(x^6 + 1)/(x^4 + 1) = A/x^2 + B/x^2 + C/x^2 + D/x^2 + E/x^2 + F/(x^2 + 1)

where A, B, C, D, E, and F are constants that need to be determined.

We can determine these constants by equating the numerator and denominator of the partial fraction representation to the original integrand and solving for the constants.

This gives us:

x^6 + 1 = A x^4 + B x^2 + C + D/x^2 + E/x^4 + F x^2 / (x^2 + 1)

By setting x = 0, we find that C = -1.

By setting x = ∞, we find that A = 0, E = 0, and F = 1.

Finally, by taking the derivative of both sides with respect to x, we find that:

6x^5 = 4Ax^3 + 2Bx + D/x^3 - 4E/x^5 + 2Fx / (x^2 + 1)

By setting x = 1, we find that:

6 = 4A + 2B + D - 4E + 2F

By substituting the values we have already determined for A, C, E, and F, we find that:

6 = 2B + D + 2

Thus, B = 2 and D = 2.

So the partial fraction decomposition is:

(x^6 + 1)/(x^4 + 1) = (2/x^2) + (2/x^2) - 1/(x^2 + 1)

We can now integrate each term of the partial fraction decomposition separately. The antiderivative of each term can be found using basic integrals:

∫ (2/x^2) dx = -2 * ln |x| + C

∫ -1/(x^2 + 1) dx = -arctan x + C

Thus, the antiderivative of the original integrand is:

∫ (x^6 + 1)/(x^4 + 1) dx = -2 * ln |x| - arctan x + C

∫2x^2+1/x^2(1+x^2)dx

这是一个比较复杂的积分，需要进行分部积分和有理函数分解。首先，我们可以将原式拆分成两个部分：

∫2x^2/x^2(1+x^2)dx + ∫1/x^2(1+x^2)dx

对于第一个积分，我们可以进行部分分式分解，得到：

2∫(1/(1+x^2))dx - 2∫(1/(1+x^2))/x^2 dx

第一个积分很容易求解，结果是 2arctan(x)。对于第二个积分，我们可以再次进行部分分式分解，得到：

2∫(1/x^2)dx - 2∫(x^2/(1+x^2))/x^2 dx

第一个积分很简单，结果是 -2/x。对于第二个积分，我们可以将分子进行展开，得到：

-2∫(1/(1+x^2))dx + 2∫(1/(1+x^2))dx/x^2 dx

第一个积分同样很容易求解，结果是 -2arctan(x)。对于第二个积分，我们可以再次进行部分分式分解，得到：

2∫(1/(1+x^2))(1/x^2 - 1/(1+x^2))dx

这个积分也可以比较容易地求解，结果是 2ln(x) - 2arctan(x)。

将这些结果合并起来，我们可以得到原式的最终结果：

2arctan(x) - 2/x - 2ln(x) + C

其中 C 是常数项。