# Calculus/Indefinite integral/Solutions

1. Evaluate $\int \frac{3x}{2}dx$

We need to find a function, $F$ , such that
$F'(x)=\frac{3x}{2}$
We know that
$\frac{d}{dx}x^{2}=2x$
So we need to find a constant, $a$ , such that
$\frac{d}{dx}ax^{2}=2ax=\frac{3x}{2}$
Solving for $a$, we get
$2ax=\frac{3x}{2}\implies a=\frac{3}{4}$
So
$\int\frac{3x}{2}=\mathbf{\frac{3}{4}x^{2}+C}$
Check your answer by taking the derivative of the function you've found and checking that it matches the integrand:
$\frac{d}{dx}(\frac{3}{4}x^{2}+C)=\frac{3}{4}(2x)=\frac{3x}{2}$

2. Find the general antiderivative of the function $f(x)=2x^4$.

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

3. Evaluate $\int(7x^2+3\cos(x)-e^x)dx$

\begin{align}\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{align}

4. Evaluate $\int(\frac{2}{5x}+\sin(x))dx$

\begin{align}\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{align}

5. Evaluate $\int x\sin(2x^2)dx$ by making the substitution $u=2x^2$

Since $u=2x^{2}$, $du=4xdx$ and $dx=\frac{du}{4x}$
\begin{align}\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{align}

6. Evaluate $\int-3\cos(x)e^{\sin(x)}dx$

Let $u=\sin(x)$, $du=\cos(x)dx$ so that $dx=\frac{du}{\cos(x)}$
\begin{align}\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{align}

7. Evaluate $\int \frac{2x-5}{x^3}dx$ using integration by parts with $u=2x-5$ and $dv=\frac{dx}{x^3}$

$du=2dx$; $v=\int\frac{dx}{x^{3}}=-\frac{1}{2x^{2}}$
\begin{align}\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{align}

8. Evaluate $\int(2x-1)e^{-3x+1}dx$

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