Evaluating $\int x^2\cosh(x)\,\mathrm dx$ $$\int x^2\cosh(x)\,\mathrm dx$$
So my working out follows -
In the first step I have made $x^2=u$, and $\cosh(x) = \dfrac{dv}{dx}$
$= uv - ∫v \dfrac{dv}{dx}$
$= x^2\sinh(x) - ∫\sinh(x)2x$
$= x^2\sinh(x) - 2x\cosh(x) - ∫2\cosh(x)$
And my final figure is $x^2\sinh(x) -2x\cosh(x) - 2\sinh(x) + c$
However wolframalpha is saying that the end term should be $+ 2\sinh(x)$ not minus. But I can't see how. 
 A: In the first step x²  = u, and cosh(x) = dv/dx
= uv - ∫v u'
= x²sinh(x) - ∫sinh(x)2x dx
= x²sinh(x) - [2xcosh(x) - ∫2cosh(x) dx]
And  final answer is x²sinh(x) -2xcosh(x) + 2sinh(x) + c
A: $\int x^{2}\cosh(x)dx=x^{2}\sinh(x)-2\int x\sinh(x)dx$
$\int x\sinh(x)dx=x\cosh(x)-\int\cosh(x)dx=x\cosh(x)-\sinh(x)+C$
Together we get:
$\int x^{2}\cosh(x)dx=x^{2}\sinh(x)-2\big(x\cosh(x)-\sinh(x)+C\big)=x^{2}\sinh(x)-2x\cosh(x)+2\sinh(x)+D$ where $D=-2C$.
A: I see your error. You only failed to distribute a negative sign. The integration being pretty short, I will recreate it. We have
$$\int x^2\cosh(x) \, dx.$$
Using the formula $\displaystyle\int u \, dv = u \, v - \int v \, du$ twice, we first take 
\begin{align*}
u=x^2 &\quad v=\sinh(x) \\
du=2x  &\quad dv=\cosh{x}.
\end{align*}
Hence the integral becomes
$$x^2\sinh(x)-2\int x \sinh(x) \, dx.$$
Applying the method again on the last integrand, we take
\begin{align*}
u=x &\quad v=\cosh(x) \\
du=1  &\quad dv=\sinh(x).
\end{align*}
Now we get
\begin{align*}
&x^2\sinh(x)-2\left[ x\cosh(x)-\int \cosh(x) \, dx \right]  \\
=&x^2\sinh(x)-2\left( x\cosh(x)-\sinh(x)\right)  \\
=&x^2\sinh(x)-2x\cosh(x)+2\sinh(x)+c.
\end{align*}
Your method was otherwise perfect.
A: \begin{align}
&\color{#ff0000}{\large\int x^{2}\cosh\left(x\right)\,{\rm d}x}
=
\left.{{\rm d}^{2} \over {\rm d}\mu^{2}}\int\cosh\left(\mu x\right)
\,{\rm d}x\right\vert_{\mu\ =\ 1}
=
\left.{{\rm d}^{2} \over {\rm d}\mu^{2}}
{\sinh\left(\mu x\right) \over \mu}\right\vert_{\mu\ =\ 1}
\\[3mm]&=
\left.{{\rm d} \over {\rm d}\mu}\left[%
x\,{\cosh\left(\mu x\right) \over \mu}
-
{\sinh\left(\mu x\right) \over \mu^{2}}
\right]\right\vert_{\mu\ =\ 1}
\\[3mm]&=
\left.\left[%
x^{2}\,{\sinh\left(\mu x\right) \over \mu}
-
x\,{\cosh\left(\mu x\right) \over \mu^{2}}
-
x\,{\cos\left(\mu x\right) \over \mu^{2}}
+
{2\sinh\left(\mu x\right) \over \mu^{3}}
\right]\right\vert_{\mu\ =\ 1}
\\[3mm]&=
\color{#ff0000}{\large\left(x^{2} + 2\right)\sinh\left(x\right)
-
2x\cosh\left(x\right)}
\end{align}
$+$ a constant.
