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Calculus Examples
∫xarccot(x)dx∫xarccot(x)dx
Step 1
Integrate by parts using the formula ∫udv=uv-∫vdu∫udv=uv−∫vdu, where u=arccot(x)u=arccot(x) and dv=xdv=x.
arccot(x)(12x2)-∫12x2(-11+x2)dxarccot(x)(12x2)−∫12x2(−11+x2)dx
Step 2
Step 2.1
Combine 1212 and x2x2.
arccot(x)x22-∫12x2(-11+x2)dxarccot(x)x22−∫12x2(−11+x2)dx
Step 2.2
Combine arccot(x)arccot(x) and x22x22.
arccot(x)x22-∫12x2(-11+x2)dxarccot(x)x22−∫12x2(−11+x2)dx
arccot(x)x22-∫12x2(-11+x2)dxarccot(x)x22−∫12x2(−11+x2)dx
Step 3
Since 12⋅-112⋅−1 is constant with respect to xx, move 12⋅-112⋅−1 out of the integral.
arccot(x)x22-(12⋅-1∫x2(11+x2)dx)arccot(x)x22−(12⋅−1∫x2(11+x2)dx)
Step 4
Step 4.1
Simplify.
Step 4.1.1
Combine 1212 and -1−1.
arccot(x)x22-(-12∫x2(11+x2)dx)arccot(x)x22−(−12∫x2(11+x2)dx)
Step 4.1.2
Move the negative in front of the fraction.
arccot(x)x22-(-12∫x2(11+x2)dx)arccot(x)x22−(−12∫x2(11+x2)dx)
Step 4.1.3
Combine x2 and 11+x2.
arccot(x)x22-(-12∫x21+x2dx)
Step 4.1.4
Multiply -1 by -1.
arccot(x)x22+1(12∫x21+x2dx)
Step 4.1.5
Multiply 12 by 1.
arccot(x)x22+12∫x21+x2dx
arccot(x)x22+12∫x21+x2dx
Step 4.2
Reorder 1 and x2.
arccot(x)x22+12∫x2x2+1dx
arccot(x)x22+12∫x2x2+1dx
Step 5
Step 5.1
Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of 0.
x2 | + | 0x | + | 1 | x2 | + | 0x | + | 0 |
Step 5.2
Divide the highest order term in the dividend x2 by the highest order term in divisor x2.
1 | |||||||||||
x2 | + | 0x | + | 1 | x2 | + | 0x | + | 0 |
Step 5.3
Multiply the new quotient term by the divisor.
1 | |||||||||||
x2 | + | 0x | + | 1 | x2 | + | 0x | + | 0 | ||
+ | x2 | + | 0 | + | 1 |
Step 5.4
The expression needs to be subtracted from the dividend, so change all the signs in x2+0+1
1 | |||||||||||
x2 | + | 0x | + | 1 | x2 | + | 0x | + | 0 | ||
- | x2 | - | 0 | - | 1 |
Step 5.5
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
1 | |||||||||||
x2 | + | 0x | + | 1 | x2 | + | 0x | + | 0 | ||
- | x2 | - | 0 | - | 1 | ||||||
- | 1 |
Step 5.6
The final answer is the quotient plus the remainder over the divisor.
arccot(x)x22+12∫1-1x2+1dx
arccot(x)x22+12∫1-1x2+1dx
Step 6
Split the single integral into multiple integrals.
arccot(x)x22+12(∫dx+∫-1x2+1dx)
Step 7
Apply the constant rule.
arccot(x)x22+12(x+C+∫-1x2+1dx)
Step 8
Since -1 is constant with respect to x, move -1 out of the integral.
arccot(x)x22+12(x+C-∫1x2+1dx)
Step 9
Step 9.1
Reorder x2 and 1.
arccot(x)x22+12(x+C-∫11+x2dx)
Step 9.2
Rewrite 1 as 12.
arccot(x)x22+12(x+C-∫112+x2dx)
arccot(x)x22+12(x+C-∫112+x2dx)
Step 10
The integral of 112+x2 with respect to x is arctan(x)+C.
arccot(x)x22+12(x+C-(arctan(x)+C))
Step 11
Simplify.
arccot(x)x22+x2-arctan(x)2+C
Step 12
Reorder terms.
12arccot(x)x2+12x-12arctan(x)+C