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Calculus Examples
∫arccot(x)dx
Step 1
Integrate by parts using the formula ∫udv=uv-∫vdu, where u=arccot(x) and dv=1.
arccot(x)x-∫x(-11+x2)dx
Step 2
Combine x and 11+x2.
arccot(x)x-∫-x1+x2dx
Step 3
Since -1 is constant with respect to x, move -1 out of the integral.
arccot(x)x--∫x1+x2dx
Step 4
Step 4.1
Multiply -1 by -1.
arccot(x)x+1∫x1+x2dx
Step 4.2
Multiply ∫x1+x2dx by 1.
arccot(x)x+∫x1+x2dx
arccot(x)x+∫x1+x2dx
Step 5
Step 5.1
Let u=1+x2. Find dudx.
Step 5.1.1
Differentiate 1+x2.
ddx[1+x2]
Step 5.1.2
By the Sum Rule, the derivative of 1+x2 with respect to x is ddx[1]+ddx[x2].
ddx[1]+ddx[x2]
Step 5.1.3
Since 1 is constant with respect to x, the derivative of 1 with respect to x is 0.
0+ddx[x2]
Step 5.1.4
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=2.
0+2x
Step 5.1.5
Add 0 and 2x.
2x
2x
Step 5.2
Rewrite the problem using u and du.
arccot(x)x+∫1u⋅12du
arccot(x)x+∫1u⋅12du
Step 6
Step 6.1
Multiply 1u by 12.
arccot(x)x+∫1u⋅2du
Step 6.2
Move 2 to the left of u.
arccot(x)x+∫12udu
arccot(x)x+∫12udu
Step 7
Since 12 is constant with respect to u, move 12 out of the integral.
arccot(x)x+12∫1udu
Step 8
The integral of 1u with respect to u is ln(|u|).
arccot(x)x+12(ln(|u|)+C)
Step 9
Simplify.
arccot(x)x+12ln(|u|)+C
Step 10
Replace all occurrences of u with 1+x2.
arccot(x)x+12ln(|1+x2|)+C