急,求不定积分In(x^2+1)dx ∫下面是0 上面是3
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![急,求不定积分In(x^2+1)dx ∫下面是0 上面是3](/uploads/image/z/12879064-64-4.jpg?t=%E6%80%A5%2C%E6%B1%82%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86In%28x%5E2%2B1%29dx+%E2%88%AB%E4%B8%8B%E9%9D%A2%E6%98%AF0+%E4%B8%8A%E9%9D%A2%E6%98%AF3)
急,求不定积分In(x^2+1)dx ∫下面是0 上面是3
急,求不定积分In(x^2+1)dx ∫下面是0 上面是3
急,求不定积分In(x^2+1)dx ∫下面是0 上面是3
∫(0,3)ln(x²+1)dx
=xln(x²+1)|(0,3)-∫(0,3)xdln(x²+1)
=3ln10-∫(0,3)2x²/(x²+1)dx
=3ln10-2∫(0,3)(x²+1-1)/(x²+1)dx
=3ln10-2∫(0,3)[1-1/(x²+1)]dx
=3ln10-2∫(0,3)dx+2∫(0,3)1/(x²+1)dx
=3ln10-6+2arctan3.
答案是: 然后你把3和0代进去相减,得3ln(10)-4+2arctan(3)。 不定积分的具体步骤: Take the integral: 你给出了上下限,这是定积分。 那个符号是ln(小写L),不是In(不是大写i)。
integral log(1+x^2) dx
For the integrand log(1+x^2), integrate by parts, integral f dg = f g- integral g df, where
f = log(1+x^2), dg = dx,
df = (2 x)/(1+x^2) dx, g = x:
= - integral (2 x^2)/(1+x^2) dx+x log(1+x^2)
Factor out constants:
= -2 integral x^2/(1+x^2) dx+x log(1+x^2)
For the integrand x^2/(1+x^2), do long division:
= -2 integral (1-1/(1+x^2)) dx+x log(1+x^2)
Integrate the sum term by term and factor out constants:
= -2 integral 1 dx+2 integral 1/(1+x^2) dx+x log(1+x^2)
The integral of 1/(1+x^2) is tan^(-1)(x):
= 2 tan^(-1)(x)-2 integral 1 dx+x log(1+x^2)
The integral of 1 is x:
= -2 x+2 tan^(-1)(x)+x log(1+x^2)+constant
Which is equal to:
Answer: |
| = 2 tan^(-1)(x)+x (-2+log(1+x^2))+constant