Note that \[\begin{eqnarray*} &\int& \;x^n\;dx&=&\frac{x^{n+1}}{n+1}+C,\;\;n \neq -1\\ &\int& \;\sin(ax)\;dx &=& -\frac{1}{a}\cos(ax)+C\\ &\int& \;e^{ax}\;dx &=& \frac{1}{a}e^{ax}+C\\ \end{eqnarray*}\]

Splitting the integral into three parts and using the above information we have:
\[\begin{eqnarray*}\simplify[std]{Int({b} * e ^ ({a}*x) + {b1} * Sin({a1}*x) + {a2} * x ^ {c3},x)}&=&\simplify[std]{Int({b} * e ^ ({a}*x),x)+Int({b1} * Sin({a1}*x),x)+Int({a2} * x ^ {c3},x) }\\ &=&\simplify[std]{({b}/{a}) * (e ^({a}*x)) + (({(-b1)}/{a1}) * Cos({a1}*x)) + ({a2}/{c3+1}) * (x ^ {(c3 + 1)})+C} \end{eqnarray*}\]

Let $y = \simplify[std]{{a}*x+{d}}$. Then,
\[\simplify[std]{{b}/(({a}*x+{d})^{n})} = \simplify[std]{{b}/(y^{n})}.\]

Now,
\[\int \simplify[std]{{b}/({a}*x+{d})^{n}} dx = \int \simplify[std]{{b}/(y^{n})} \frac{dx}{dy} dy.\]

Rearrange $y = \simplify[std]{{a}x+{d}}$ to get $\displaystyle x = \simplify[std]{(y-{b})/{a}}$, and hence $\displaystyle\frac{dx}{dy} = \frac{1}{\var{a}}$.

$\displaystyle \int \frac{1}{y^n} dx = -\frac{1}{(n-1)y^{n-1}} + C$ is a standard integral, so we can now calculate the desired integral:

\[\int \simplify[std]{{b}/(y^{n})} \frac{dx}{dy} dy = \simplify[std]{{b}/({n-1}*y^{n-1})} \cdot \frac{1}{\var{a}} + C = \simplify[std]{(-{b})/({a*(n-1)}*({a}*x+{d})^{n-1}) + C}.\]

Let $y = \simplify[std]{{a}*x+{d}}$.

Then $\displaystyle x=\frac{1}{\var{a}}\simplify[std]{(y-{d})}$ and so we have the numerator $\simplify[std]{{b}*x+{c}}$ becomes in terms of $y$:

$\displaystyle \simplify[std]{{b}*x+{c} = {b}*1/{a}*(y-{d})+{c}= {m}y+{r}}$ and so

\[\displaystyle \simplify[std]{({b}*x+{c})/(({a}*x+{d})^{n})} = \simplify[std]{({m}*y+{r})/(y^{n})={m}/y^{n-1}+{r}/y^{n}}\]

Now,
\[\int \simplify[std]{({b}x+{c})/({a}*x+{d})^{n}} dx = \int \left(\simplify[std]{{m}/y^{n-1}+{r}/y^{n}} \right)\frac{dx}{dy} dy \]

Since $\displaystyle x = \simplify[std]{(y-{d})/{a}}$ then $\displaystyle \frac{dx}{dy} = \frac{1}{\var{a}}$.

We can now calculate the desired integral:

\[ \begin{eqnarray*} \int \left(\simplify[std]{{m}/y^{n-1}+{r}/y^{n}}\right) \frac{dx}{dy} dy &=&\frac{1}{\var{a}}\left(\int \simplify[std]{{m}/y^{n-1}}\;dy+\int \simplify[std]{{r}/y^{n}}\;dy \right)\\ &=&\frac{1}{\var{a}}\left(\simplify[std]{{-m}/({n-2}*y^{n-2})+ {-r}/({n-1}*y^{n-1})}\right) + C \\ &=& \simplify[std]{(-{m})/({a*(n-2)}*({a}*x+{d})^{n-2})+(-{r})/({a*(n-1)}*({a}*x+{d})^{n-1}) + C}\\ &=&\simplify[std]{1/({a}x+{d})^{n-1}*(({-m}/{a*(n-2)})*({a}x+{d})-{r}/({a*(n-1)}))}\\ &=&\simplify[std]{1/({a}x+{d})^{n-1}*(({-m}/{(n-2)})*x+{-m*d*(n-1)-r*(n-2)}/{(n-2)*(n-1)*a})} \end{eqnarray*} \]
Hence \[g(x)=\simplify[std]{({-m}/{(n-2)})*x+{-m*d*(n-1)-r*(n-2)}/{(n-2)*(n-1)*a}}\]

a)

The formula for integrating by parts is
\[ \int u\frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx. \]

We choose $u = \simplify[std]{({a}x+{b})}$ and $\displaystyle \frac{dv}{dx} = \simplify[std]{cos({c}*x+{d})}$.

So $\displaystyle \frac{du}{dx} = \simplify[std]{{a}}$ and $\displaystyle v = \simplify[std]{(1/{c})*sin({c}*x+{d})}$.

Hence,
\[ \begin{eqnarray} \int \simplify[std]{({a}*x+{b})*cos({c}*x+{d})} dx &=& uv - \int v \frac{du}{dx} dx \\ &=& \simplify[std]{(({a}*x+{b})/{c})*sin({c}*x+{d}) - ({a}/{c})*Int(sin({c}*x+{d}),x)} \\ &=& \simplify[std]{(({a}x+{b})/{c})*sin({c}*x+{d}) +({a}/{c^2})*cos({c}*x+{d}) + C} \end{eqnarray} \]

b)

For this part we choose $u = \simplify[std]{({a}x+{b})^2}$ and $\displaystyle \frac{dv}{dx} = \simplify[std]{sin({c}*x+{d})}$.

So $\displaystyle \frac{du}{dx}=\simplify[std]{{2*a}*({a}*(x)+{b})}$ and $\displaystyle v = \simplify[std]{-(1/{c})*cos({c}*x+{d})}$.

Hence,
\[ \begin{eqnarray*}I= \int \simplify[std]{({a}*x+{b})^2*e^({c}*x)} dx &=& uv - \int v \frac{du}{dx} dx \\ &=& \simplify[std]{({-1}/{c})*({a}x+{b})^2*cos({c}*x+{d}) + (1/{c})*Int({2*a}*({a}x+{b})*cos({c}*x+{d}),x)} \\ &=& \simplify[std]{({-1}/{c})*({a}x+{b})^2*cos({c}*x+{d}) +({2*a}/{c})*Int(({a}x+{b})*cos({c}*x+{d}),x)}\dots (*) \end{eqnarray*}\]

But in Part a we have aready worked out $\displaystyle \simplify[std]{Int(({a}x+{b})*cos({c}*x+{d}),x)=(({a}x+{b})/{c})*sin({c}*x+{d}) +({a}/{c^2})*cos({c}*x+{d})}$ 

So on substituting this in equation (*) we find:
\[ \begin{eqnarray*}I&=& \simplify[std]{({-1}/{c})*({a}x+{b})^2*cos({c}*x+{d}) +({2*a}/{c})*((({a}x+{b})/{c})*sin({c}*x+{d}) +({a}/{c^2})*cos({c}*x+{d}))+C}\\ &=& \simplify[std]{-(({a}*x+{b})^2/{c})*cos({c}*x+{d})+(({2*a}({a}x+{b}))/{c^2})*sin({c}*x+{d})+({2*a^2}/{c^3})*cos({c}*x+{d})+C} \end{eqnarray*}\]

c)

Let $\displaystyle A= \simplify[std]{int(exp({c1}x)*( {u}*sin({d1}x)+{1-u}*cos({d1}x)),x)} $. We solve this using two integration by parts, and we choose $u = \simplify[std]{ {u}*sin({d1}x)+{1-u}*cos({d1}x)}$ in both. 

\[\begin{eqnarray*} A&=&\simplify[std]{ 1/{c1}exp({c1}x)*( {u}*sin({d1}x)+{1-u}*cos({d1}x))+{((-1)^u)*d1}/{c1}int(exp({c1}x) *({u}*cos({d1}x)+{(1-u)}*sin({d1}x)),x)}\\&=&\simplify[std]{1/{c1}exp({c1}x)*( {u}*sin({d1}x)+{1-u}*cos({d1}x))+{((-1)^u)*d1}/{c1}*(1/{c1}exp({c1}x)*( {u}*cos({d1}x)+{1-u}*sin({d1}x))+{(-1)^(u+1)*d1}/{c1}int(exp({c1}x)*( {u}*sin({d1}x)+{1-u}*cos({d1}x)),x) )}\\&=&\simplify[std]{1/{c1}exp({c1}x)*( {u}*sin({d1}x)+{1-u}*cos({d1}x))+{((-1)^u)*d1}/{c1}*(1/{c1}exp({c1}x)*( {u}*cos({d1}x)+{1-u}*sin({d1}x))+{(-1)^(u+1)*d1}/{c1}A )}\end{eqnarray*}\]

Note that after integrating by parts twice, we have the integral $A$ on both sides of this equation.

Rearranging we have: \[A = \simplify[std]{e^({c1}x)/{c1^2+d1^2}*(({u*(c1-d1)+d1})*sin({d1}x)+({u*(-c1-d1)+c1})*cos({d1}x))+C}\]