Question

てつ

途中式をお願いします！！

1. 次の不定積分を求めよ.
.3
(1) √√√₂+³+1
S
dx
x sin¹ x
√1-x²
dx
(3) S
(6) S
S
dx
5x+3/ x2
1-ex
(8) S (170)2 dx
(1+ex)²
1
(2) Soorty dx
cos x
x² +1
(x-1)³
1
(7) S2-tan² x
(4) S S
dx (tan x = t)
dx
X³
(5) √ √(x² + 1)²
dx (tan x = t)
dx

-1
4
1. (1) 1 tan ¹x²
4
(2)
(3) x-√1-x2² sin¯¹ x
1
3
tan³ x+tan x
(4) log |x-1|–
2x-1
(x-1)²

1
(5) 2(x²+1) + ²log (x²+1)
2
1
1
(7) = (x + 2√2 log |√2+tan x)
3
(8) log (₁) +1+²+
(1
ex
2
e*
e*
(6)
3|5
log 5x+1

解析学、積分、不定積分