Find the solution of {f}{''}{\left({x}\right)}={8}{x}+ \sin{{x}}

Solve ${\displaystyle \frac{d^{2}f\left(x\right)}{{dx}^2}=8x+\sin \left(x\right):}$
Integrate both sides with respect to x:
${\displaystyle \frac{df\left(x\right)}{dx}=\int (8x+\sin \left(x\right))dx=4x^2-\cos \left(x\right)+c_1}$, where ${\displaystyle c_1}$ is an arbitrary constant.
INTERMEDIATE STEPS:
Take the integral:
${\displaystyle \int (8x+\sin \left(x\right))dx}$
Integrate the sum term by term and factor out constants:
${\displaystyle =\int \sin \left(x\right)dx+8\int xdx}$
The integral of $\sin (x)$ is $-\cos (x)$:
${\displaystyle =-\cos \left(x\right)+8\int xdx}$
The integral of x is ${\displaystyle \frac{x^2}{2}:}$
${\displaystyle =4x^2-\cos \left(x\right)+\text{constant}}$
Integrate both sides with respect to x:
${\displaystyle f\left(x\right)=\int (4x^2-\cos \left(x\right)+c_1)dx=\frac{4x^3}{3}-\sin \left(x\right)+x{c}_1+c_2}$, where ${\displaystyle c_2}$ is an arbitrary constant.
INTERMEDIATE STEPS:
Take the integral:
${\displaystyle \int (c_1+4x^2-\cos \left(x\right))dx}$
Integrate the sum term by term and factor out constants:
${\displaystyle =c_1\int 1dx-\int \cos \left(x\right)dx+4\int x^{2}dx}$
The integral of 1 is x:
${\displaystyle =c_{1}x-\int \cos \left(x\right)dx+4\int x^{2}dx}$
The integral of $\cos (x)$ is $\sin (x)$:
${\displaystyle =-\sin \left(x\right)+c_{1}x+4\int x^{2}dx}$
The integral of ${\displaystyle x^2}$ is ${\displaystyle \frac{x^3}{3}}$:
Answer:
${\displaystyle =c_{1}x+\frac{4x^3}{3}-\sin \left(x\right)+\text{constant}}$