# How do you integrate 8x(x²+1)³ dx ?

Sep 16, 2016

${\left({x}^{2} + 1\right)}^{4} + C$.

#### Explanation:

Let $I = \int 8 x {\left({x}^{2} + 1\right)}^{3} \mathrm{dx}$

We use the Method of Substn.

We subst. $\left({x}^{2} + 1\right) = y , \text{ so that, } 2 x \mathrm{dx} = \mathrm{dy}$. Hence,

$I = 4 \int {y}^{3} \mathrm{dy} = 4 \cdot {y}^{4} / 4 = {y}^{4} = {\left({x}^{2} + 1\right)}^{4} + C$.

Alternatively, without substn., we can find $I$.

Using ${\left(a + b\right)}^{3} = {a}^{3} + {b}^{3} + 3 a b \left(a + b\right)$, we expand ${\left({x}^{2} + 1\right)}^{3}$.

:. I=int{8x(x^6+1+3*x^2*1(x^2+1)}dx

$= 8 \int \left({x}^{7} + x + 3 {x}^{5} + 3 {x}^{3}\right) \mathrm{dx}$

$= 8 \left\{{x}^{8} / 8 + {x}^{2} / 2 + 3 \cdot {x}^{6} / 6 + 3 \cdot {x}^{4} / 4\right\}$

$= {x}^{8} + 4 {x}^{2} + 4 {x}^{6} + 6 {x}^{4} = {x}^{8} + 4 {x}^{6} + 6 {x}^{4} + 4 {x}^{2} + K$.

On the first hand, these two Answers may look different, but,

bearing in mind the binomial expansion of

${\left(p + 1\right)}^{4} = {p}^{4} + 4 {p}^{3} + 6 {p}^{2} + 4 p + 1$, we find that,

$I = {x}^{8} + 4 {x}^{6} + 6 {x}^{4} + 4 {x}^{2} + K$,

$= {x}^{8} + 4 {x}^{6} + 6 {x}^{4} + 4 {x}^{2} + 1 + C '$, where, $C ' = K - 1$

$I = {\left({x}^{2} + 1\right)}^{4} + C ' , C ' = K - 1$

Enjoy Maths.!