Find the indefinite integral. ∫x(x2+1)2dx

Question

Find the indefinite integral.  ∫x(x2+1)2dx

Oung1985 · Accepted Answer

Step 1
We have to find the indefinite integral:

\int x {(x^{2} + 1)}^{2} d x

We will solve this integrals using substitution method:
Let

t = x^{2} + 1

Differentiating with respect to x, we get

t = x^{2} + 1

dt=2xdx+0

\frac{d t}{2} = x d x

So putting the value of xdx in the given indefinite integration, we get
Step 2
Here,

\int x {(x^{2} + 1)}^{2} d x = \int {(x^{2} + 1)}^{2} x d x

= \int t^{2} (\frac{d t}{2})

= \frac{1}{2} \int t^{2} d t

= \frac{1}{2} (\frac{t^{2 + 1}}{2 + 1}) + c

= \frac{t^{3}}{6} + c

.
(using formula

\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c

)
Where c is an arbitrary constant.
Putting

t = x^{2} + 1

, we get

= \frac{{(x^{2} + 1)}^{3}}{6} + c

.
Hence, value of indefinite integration is

\frac{{(x^{2} + 1)}^{3}}{6} + c

.

Daniel Williams · Accepted Answer

∫x(x2+1)2dx=x66+x42+x22+C  Explanation:  Expand (x2+1)2  (x2+1)2=x4+2x2+1  Distribute the x  x(x4+2x2+1)=x5+2x3+x  Next we integrate each term  ∫x5+2x3+xdx=∫x5dx+∫2x3dx+∫xdx  =x66+2⋅x44+x22  =x66+x42+x22+C

Find the indefinite integral. \int x(x^{2}+1)^{2}dx

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