Integral of $$$\frac{\sin{\left(x \right)}}{x + 1}$$$

Find $$$\int \frac{\sin{\left(x \right)}}{x + 1}\, dx$$$.

Let $$$u=x + 1$$$.

Then $$$du=\left(x + 1\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

So,

$${\color{red}{\int{\frac{\sin{\left(x \right)}}{x + 1} d x}}} = {\color{red}{\int{\frac{\sin{\left(u - 1 \right)}}{u} d u}}}$$

Rewrite the integrand:

$${\color{red}{\int{\frac{\sin{\left(u - 1 \right)}}{u} d u}}} = {\color{red}{\int{\frac{\sin{\left(u \right)} \cos{\left(1 \right)} - \sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u}}}$$

Expand the expression:

$${\color{red}{\int{\frac{\sin{\left(u \right)} \cos{\left(1 \right)} - \sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u}}} = {\color{red}{\int{\left(\frac{\sin{\left(u \right)} \cos{\left(1 \right)}}{u} - \frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u}\right)d u}}}$$

Integrate term by term:

$${\color{red}{\int{\left(\frac{\sin{\left(u \right)} \cos{\left(1 \right)}}{u} - \frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u}\right)d u}}} = {\color{red}{\left(- \int{\frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u} + \int{\frac{\sin{\left(u \right)} \cos{\left(1 \right)}}{u} d u}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\cos{\left(1 \right)}$$$ and $$$f{\left(u \right)} = \frac{\sin{\left(u \right)}}{u}$$$:

$$- \int{\frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u} + {\color{red}{\int{\frac{\sin{\left(u \right)} \cos{\left(1 \right)}}{u} d u}}} = - \int{\frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u} + {\color{red}{\cos{\left(1 \right)} \int{\frac{\sin{\left(u \right)}}{u} d u}}}$$

This integral (Sine Integral) does not have a closed form:

$$- \int{\frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u} + \cos{\left(1 \right)} {\color{red}{\int{\frac{\sin{\left(u \right)}}{u} d u}}} = - \int{\frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u} + \cos{\left(1 \right)} {\color{red}{\operatorname{Si}{\left(u \right)}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\sin{\left(1 \right)}$$$ and $$$f{\left(u \right)} = \frac{\cos{\left(u \right)}}{u}$$$:

$$\cos{\left(1 \right)} \operatorname{Si}{\left(u \right)} - {\color{red}{\int{\frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u}}} = \cos{\left(1 \right)} \operatorname{Si}{\left(u \right)} - {\color{red}{\sin{\left(1 \right)} \int{\frac{\cos{\left(u \right)}}{u} d u}}}$$

This integral (Cosine Integral) does not have a closed form:

$$\cos{\left(1 \right)} \operatorname{Si}{\left(u \right)} - \sin{\left(1 \right)} {\color{red}{\int{\frac{\cos{\left(u \right)}}{u} d u}}} = \cos{\left(1 \right)} \operatorname{Si}{\left(u \right)} - \sin{\left(1 \right)} {\color{red}{\operatorname{Ci}{\left(u \right)}}}$$

Recall that $$$u=x + 1$$$:

$$- \sin{\left(1 \right)} \operatorname{Ci}{\left({\color{red}{u}} \right)} + \cos{\left(1 \right)} \operatorname{Si}{\left({\color{red}{u}} \right)} = - \sin{\left(1 \right)} \operatorname{Ci}{\left({\color{red}{\left(x + 1\right)}} \right)} + \cos{\left(1 \right)} \operatorname{Si}{\left({\color{red}{\left(x + 1\right)}} \right)}$$

Therefore,

$$\int{\frac{\sin{\left(x \right)}}{x + 1} d x} = - \sin{\left(1 \right)} \operatorname{Ci}{\left(x + 1 \right)} + \cos{\left(1 \right)} \operatorname{Si}{\left(x + 1 \right)}$$

Add the constant of integration:

$$\int{\frac{\sin{\left(x \right)}}{x + 1} d x} = - \sin{\left(1 \right)} \operatorname{Ci}{\left(x + 1 \right)} + \cos{\left(1 \right)} \operatorname{Si}{\left(x + 1 \right)}+C$$

$$$\int \frac{\sin{\left(x \right)}}{x + 1}\, dx = \left(- \sin{\left(1 \right)} \operatorname{Ci}{\left(x + 1 \right)} + \cos{\left(1 \right)} \operatorname{Si}{\left(x + 1 \right)}\right) + C$$$A