Question

Solve the integral:$\int \frac{3x+1}{secx}dx$

Short answer

The required answer is$\left(3x+1\right)\sin x+3\cos x+c$.

Step-by-step solution

Step 1. Given information. 

We have given integral is$\int \frac{3x+1}{secx}dx$.

Step 2. Solve the integration by parts .  

We have,

$$\begin{gathered} u=3x+1 \\ du=3dx \end{gathered}$$

and

$$\begin{gathered} dv=\cos xdx \\ v=\int \cos xdx \\ v=\sin x \end{gathered}$$

The formula of integration by parts is $\int udv=uv-\int vdu$.

localid="1648742547412" $$\begin{gathered} \int \frac{3x+1}{secx}dx \\ =(3x+1)\sin x-\int 3\sin xdx \\ =(3x+1)\sin x-3\int \sin xdx \\ =(3x+1)\sin x-3(-\cos x)+c \\ =(3x+1)\sin x+3\cos x+c \end{gathered}$$