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Find the integral $\int\left(z^2+3z\right)\arctan\left(z\right)dz$

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Solution

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Final answer to the problem

$\frac{z^{3}\arctan\left(z\right)}{3}+\frac{3}{2}z^2\arctan\left(z\right)-\frac{3}{2}z+\frac{3}{2}\arctan\left(z\right)-\frac{1}{6}z^2+\frac{1}{6}\ln\left|1+z^2\right|+C_0$
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Step-by-step Solution

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  • Integrate by partial fractions
  • Integrate by substitution
  • Integrate by parts
  • Integrate using tabular integration
  • Integrate by trigonometric substitution
  • Weierstrass Substitution
  • Integrate using trigonometric identities
  • Integrate using basic integrals
  • Product of Binomials with Common Term
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We can solve the integral $\int\left(z^2+3z\right)\arctan\left(z\right)dz$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula

Learn how to solve integral calculus problems step by step online.

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$

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Learn how to solve integral calculus problems step by step online. Find the integral int((z^2+3z)arctan(z))dz. We can solve the integral \int\left(z^2+3z\right)\arctan\left(z\right)dz by applying integration by parts method to calculate the integral of the product of two functions, using the following formula. First, identify or choose u and calculate it's derivative, du. Now, identify dv and calculate v. Solve the integral to find v.

Final answer to the problem

$\frac{z^{3}\arctan\left(z\right)}{3}+\frac{3}{2}z^2\arctan\left(z\right)-\frac{3}{2}z+\frac{3}{2}\arctan\left(z\right)-\frac{1}{6}z^2+\frac{1}{6}\ln\left|1+z^2\right|+C_0$

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Function Plot

Plotting: $\frac{z^{3}\arctan\left(z\right)}{3}+\frac{3}{2}z^2\arctan\left(z\right)-\frac{3}{2}z+\frac{3}{2}\arctan\left(z\right)-\frac{1}{6}z^2+\frac{1}{6}\ln\left(1+z^2\right)+C_0$

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7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
◻2
◻◻
√◻
√

◻√◻
◻√
∞
e
π
ln
log
log◻
lim
d/dx
D□x
∫
∫◻
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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Main Topic: Integral Calculus

Integration assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise by combining infinitesimal data.

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