作业帮急,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
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