# Calculus/Indefinite integral/Solutions

1. Evaluate ${\displaystyle \int {\frac {3x}{2}}dx}$
We need to find a function, ${\displaystyle F}$ , such that
${\displaystyle F'(x)={\frac {3x}{2}}}$

We know that

${\displaystyle {\frac {d}{dx}}x^{2}=2x}$

So we need to find a constant, ${\displaystyle a}$ , such that

${\displaystyle {\frac {d}{dx}}ax^{2}=2ax={\frac {3x}{2}}}$

Solving for ${\displaystyle a}$ , we get

${\displaystyle 2ax={\frac {3x}{2}}\implies a={\frac {3}{4}}}$

So

${\displaystyle \int {\frac {3x}{2}}={\frac {3}{4}}x^{2}+C}$

Check your answer by taking the derivative of the function you've found and checking that it matches the integrand:

${\displaystyle {\frac {d}{dx}}\left({\frac {3}{4}}x^{2}+C\right)={\frac {3}{4}}(2x)={\frac {3x}{2}}}$
We need to find a function, ${\displaystyle F}$ , such that
${\displaystyle F'(x)={\frac {3x}{2}}}$

We know that

${\displaystyle {\frac {d}{dx}}x^{2}=2x}$

So we need to find a constant, ${\displaystyle a}$ , such that

${\displaystyle {\frac {d}{dx}}ax^{2}=2ax={\frac {3x}{2}}}$

Solving for ${\displaystyle a}$ , we get

${\displaystyle 2ax={\frac {3x}{2}}\implies a={\frac {3}{4}}}$

So

${\displaystyle \int {\frac {3x}{2}}={\frac {3}{4}}x^{2}+C}$

Check your answer by taking the derivative of the function you've found and checking that it matches the integrand:

${\displaystyle {\frac {d}{dx}}\left({\frac {3}{4}}x^{2}+C\right)={\frac {3}{4}}(2x)={\frac {3x}{2}}}$
2. Find the general antiderivative of the function ${\displaystyle f(x)=2x^{4}}$.
We know that

${\displaystyle {\frac {d}{dx}}x^{5}=5x^{4}}$
We need to find a constant, ${\displaystyle a}$, such that
${\displaystyle {\frac {d}{dx}}ax^{5}=5ax^{4}=2x^{4}}$
Solving for ${\displaystyle a}$, we get
${\displaystyle 5ax^{4}=2x^{4}\implies a={\frac {2}{5}}}$
So the general antiderivative will be
${\displaystyle \mathbf {{\frac {2}{5}}x^{5}+C} }$
Check your answer by taking the derivative of the antiderivative you've found and checking that you get back the function you started with:

${\displaystyle {\frac {d}{dx}}\int 2x^{4}dx={\frac {d}{dx}}({\frac {2}{5}}x^{5}+C)={\frac {2}{5}}(5x^{4})=2x^{4}}$
We know that

${\displaystyle {\frac {d}{dx}}x^{5}=5x^{4}}$
We need to find a constant, ${\displaystyle a}$, such that
${\displaystyle {\frac {d}{dx}}ax^{5}=5ax^{4}=2x^{4}}$
Solving for ${\displaystyle a}$, we get
${\displaystyle 5ax^{4}=2x^{4}\implies a={\frac {2}{5}}}$
So the general antiderivative will be
${\displaystyle \mathbf {{\frac {2}{5}}x^{5}+C} }$
Check your answer by taking the derivative of the antiderivative you've found and checking that you get back the function you started with:

${\displaystyle {\frac {d}{dx}}\int 2x^{4}dx={\frac {d}{dx}}({\frac {2}{5}}x^{5}+C)={\frac {2}{5}}(5x^{4})=2x^{4}}$
3. Evaluate ${\displaystyle \int (7x^{2}+3\cos(x)-e^{x})dx}$
{\displaystyle {\begin{aligned}\int (7x^{2}+3\cos(x)-e^{x})dx&=7\int x^{2}dx+3\int \cos(x)dx-\int e^{x}dx\\&=7({\frac {x^{3}}{3}})+3\sin(x)-e^{x}+C\\&=\mathbf {{\frac {7}{3}}x^{3}+3\sin(x)-e^{x}+C} \end{aligned}}}
{\displaystyle {\begin{aligned}\int (7x^{2}+3\cos(x)-e^{x})dx&=7\int x^{2}dx+3\int \cos(x)dx-\int e^{x}dx\\&=7({\frac {x^{3}}{3}})+3\sin(x)-e^{x}+C\\&=\mathbf {{\frac {7}{3}}x^{3}+3\sin(x)-e^{x}+C} \end{aligned}}}
4. Evaluate ${\displaystyle \int ({\frac {2}{5x}}+\sin(x))dx}$
{\displaystyle {\begin{aligned}\int ({\frac {2}{5x}}+\sin(x))dx&={\frac {2}{5}}\int {\frac {dx}{x}}+\int \sin(x)dx\\&=\mathbf {{\frac {2}{5}}\ln |x|-\cos(x)+C} \end{aligned}}}
{\displaystyle {\begin{aligned}\int ({\frac {2}{5x}}+\sin(x))dx&={\frac {2}{5}}\int {\frac {dx}{x}}+\int \sin(x)dx\\&=\mathbf {{\frac {2}{5}}\ln |x|-\cos(x)+C} \end{aligned}}}
5. Evaluate ${\displaystyle \int x\sin(2x^{2})dx}$ by making the substitution ${\displaystyle u=2x^{2}}$
Since ${\displaystyle u=2x^{2}}$, ${\displaystyle du=4xdx}$ and ${\displaystyle dx={\frac {du}{4x}}}$
{\displaystyle {\begin{aligned}\int x\sin(2x^{2})dx&=\int x\sin(u){\frac {du}{4x}}\\&={\frac {1}{4}}\int \sin(u)du\\&=-{\frac {\cos(u)}{4}}+C\\&=-\mathbf {{\frac {\cos(2x^{2})}{4}}+C} \end{aligned}}}
Since ${\displaystyle u=2x^{2}}$, ${\displaystyle du=4xdx}$ and ${\displaystyle dx={\frac {du}{4x}}}$
{\displaystyle {\begin{aligned}\int x\sin(2x^{2})dx&=\int x\sin(u){\frac {du}{4x}}\\&={\frac {1}{4}}\int \sin(u)du\\&=-{\frac {\cos(u)}{4}}+C\\&=-\mathbf {{\frac {\cos(2x^{2})}{4}}+C} \end{aligned}}}
6. Evaluate ${\displaystyle \int -3\cos(x)e^{\sin(x)}dx}$
Let ${\displaystyle u=\sin(x)}$, ${\displaystyle du=\cos(x)dx}$ so that ${\displaystyle dx={\frac {du}{\cos(x)}}}$
{\displaystyle {\begin{aligned}\int -3\cos(x)e^{\sin(x)}dx&=-3\int \cos(x)e^{u}{\frac {du}{\cos(x)}}\\&=-3\int e^{u}du\\&=-3e^{u}+C\\&=\mathbf {-3e^{\sin(x)}+C} \end{aligned}}}
Let ${\displaystyle u=\sin(x)}$, ${\displaystyle du=\cos(x)dx}$ so that ${\displaystyle dx={\frac {du}{\cos(x)}}}$
{\displaystyle {\begin{aligned}\int -3\cos(x)e^{\sin(x)}dx&=-3\int \cos(x)e^{u}{\frac {du}{\cos(x)}}\\&=-3\int e^{u}du\\&=-3e^{u}+C\\&=\mathbf {-3e^{\sin(x)}+C} \end{aligned}}}
7. Evaluate ${\displaystyle \int {\frac {2x-5}{x^{3}}}dx}$ using integration by parts with ${\displaystyle u=2x-5}$ and ${\displaystyle dv={\frac {dx}{x^{3}}}}$
${\displaystyle du=2dx}$; ${\displaystyle v=\int {\frac {dx}{x^{3}}}=-{\frac {1}{2x^{2}}}}$
{\displaystyle {\begin{aligned}\int {\frac {2x-5}{x^{3}}}dx&=\int udv\\&=uv-\int vdu\\&=(2x-5)(-{\frac {1}{2x^{2}}})-\int (-{\frac {1}{2x^{2}}})2dx\\&={\frac {5-2x}{2x^{2}}}+\int {\frac {dx}{x^{2}}}\\&={\frac {5-2x}{2x^{2}}}-{\frac {1}{x}}\\&={\frac {5-2x}{2x^{2}}}-{\frac {2x}{2x^{2}}}\\&=\mathbf {\frac {5-4x}{2x^{2}}} \end{aligned}}}
${\displaystyle du=2dx}$; ${\displaystyle v=\int {\frac {dx}{x^{3}}}=-{\frac {1}{2x^{2}}}}$
{\displaystyle {\begin{aligned}\int {\frac {2x-5}{x^{3}}}dx&=\int udv\\&=uv-\int vdu\\&=(2x-5)(-{\frac {1}{2x^{2}}})-\int (-{\frac {1}{2x^{2}}})2dx\\&={\frac {5-2x}{2x^{2}}}+\int {\frac {dx}{x^{2}}}\\&={\frac {5-2x}{2x^{2}}}-{\frac {1}{x}}\\&={\frac {5-2x}{2x^{2}}}-{\frac {2x}{2x^{2}}}\\&=\mathbf {\frac {5-4x}{2x^{2}}} \end{aligned}}}
8. Evaluate ${\displaystyle \int (2x-1)e^{-3x+1}dx}$
Let ${\displaystyle u=2x-1}$; ${\displaystyle dv=e^{-3x+1}dx}$

Then ${\displaystyle du=2dx}$ and ${\displaystyle v=\int e^{-3x+1}dx}$
To evaluate ${\displaystyle v}$, make the substitution ${\displaystyle w=-3x+1}$; ${\displaystyle dw=-3dx}$; ${\displaystyle dx={\frac {-dw}{3}}}$. Then
${\displaystyle v=\int e^{-3x+1}dx=\int e^{w}({\frac {-1}{3}})dw={\frac {-e^{w}}{3}}={\frac {-e^{-3x+1}}{3}}}$. So

{\displaystyle {\begin{aligned}\int (2x-1)e^{-3x+1}dx&=\int udv\\&=uv-\int vdu\\&=(2x-1){\frac {-e^{-3x+1}}{3}}-\int {\frac {-e^{-3x+1}}{3}}(2)dx\\&={\frac {(1-2x)e^{-3x+1}}{3}}+{\frac {2}{3}}\int e^{-3x+1}dx\\&={\frac {(1-2x)e^{-3x+1}}{3}}+{\frac {2}{3}}\int {\frac {-e^{w}}{3}}dw\\&={\frac {3(1-2x)e^{-3x+1}}{9}}-{\frac {2}{9}}e^{w}\\&={\frac {(3-6x)e^{-3x+1}}{9}}-{\frac {2}{9}}e^{-3x+1}\\&=\mathbf {\frac {(1-6x)e^{-3x+1}}{9}} \end{aligned}}}
Let ${\displaystyle u=2x-1}$; ${\displaystyle dv=e^{-3x+1}dx}$

Then ${\displaystyle du=2dx}$ and ${\displaystyle v=\int e^{-3x+1}dx}$
To evaluate ${\displaystyle v}$, make the substitution ${\displaystyle w=-3x+1}$; ${\displaystyle dw=-3dx}$; ${\displaystyle dx={\frac {-dw}{3}}}$. Then
${\displaystyle v=\int e^{-3x+1}dx=\int e^{w}({\frac {-1}{3}})dw={\frac {-e^{w}}{3}}={\frac {-e^{-3x+1}}{3}}}$. So

{\displaystyle {\begin{aligned}\int (2x-1)e^{-3x+1}dx&=\int udv\\&=uv-\int vdu\\&=(2x-1){\frac {-e^{-3x+1}}{3}}-\int {\frac {-e^{-3x+1}}{3}}(2)dx\\&={\frac {(1-2x)e^{-3x+1}}{3}}+{\frac {2}{3}}\int e^{-3x+1}dx\\&={\frac {(1-2x)e^{-3x+1}}{3}}+{\frac {2}{3}}\int {\frac {-e^{w}}{3}}dw\\&={\frac {3(1-2x)e^{-3x+1}}{9}}-{\frac {2}{9}}e^{w}\\&={\frac {(3-6x)e^{-3x+1}}{9}}-{\frac {2}{9}}e^{-3x+1}\\&=\mathbf {\frac {(1-6x)e^{-3x+1}}{9}} \end{aligned}}}