calculus - Evaluate the indefinite integral

$ I = \int (x^2 + 2x)\cos(x) dx $

Integration by Parts, choose $u$:

$$\begin{align}
u &= \cos(x) \\

dv &= (x^2 + 2x)dx \\
du &= -\sin(x) \\
v &= \frac{1}{3}x^3 + x^2
\end{align}
$$

Substitute into formula:
$$
\begin{align}
\int udu &= uv - \int vdu \\

&= \cos(x)\left(\frac{1}{3}x^3 + x^2\right) - \int\left(\frac{1}{3}x^3 + 2x\right)(-\sin(x)) \\
&= \cos(x)\left(\frac{1}{3}x^3 + x^2\right) + \int\left(\frac{1}{3}x^3 + 2x\right)(\sin(x))
\end{align}
$$

At this point, it doesn't look like I can use the substitution rule on the the right hand integral, so I decide to use the substitution rule again.

Integration by Parts II, choose $u$:

$$\begin{align}

u &= sinx \\
dv &= (\frac{1}{3}x^3 + 2x)dx \\
du &= cosx \\
v &= \frac{1}{12}x^4 + x^2
\end{align}
$$

Substitute into formula:
$$\begin{align}
\int_{}udu &= uv - \int_{}vdu \\

&= (sinx)(\frac{1}{12}x^4 + x^2) - \int_{} (\frac{1}{12}x^4 + x^2)(cosx)dx
\end{align}
$$

Combining the two integration by parts together and I feel like I am no closer to evaluating the integral than whence I started...The integral is still there and I feel another parts by integration won't work.
$$\int(x^2 + 2x)\cos(x) = (\cos(x))\left(\frac{1}{3}x^3 + x^2\right) + (\sin(x))\left(\frac{1}{12}x^4 + x^2\right) - \int\left(\frac{1}{12}x^4 + x^2\right)(\cos(x))dx$$

Did I do the math wrong and make a mistake somewhere? Or am I supposed to approach this differently?

Answer

I would do this way
$$\int(x^2+2x)\cos x\ dx$$

By Integration By Parts: $u=(x^2+2x),v^{\prime}=\cos x$
$$=(x^2+2x)\sin x-\int(2x+2)\sin xdx$$

Again apply Integration By Parts for $\int(2x+2)\sin xdx$ $u=(2x+2), v^{\prime}=\sin x$
and we get

$\int(2x+2)\sin xdx=2\sin x-\cos x(2x+2)$

So, we finally get,

$$=(x^2+2x)\sin x-[2\sin x-\cos x(2x+2)]+C$$
$$=(x^2+2x-2)\sin x+2(x+1)+C$$

$$\int(x^2+2x)\cos x\ dx=(x^2+2x-2)\sin x+2(x+1)+C$$