Solve f 1 / x + 1 d t = ∫ (from 1 to frac { 1) of { x } } 1/1+t^2dt

$$f \frac { 1 } { x + 1 } d t = \int _ { 1 } ^ { \frac { 1 } { x } } \frac { 1 } { 1 + t ^ { 2 } } d t$$

Solve for d

$$\left\{\begin{matrix}d=-\frac{\left(x+1\right)\left(-4\arctan(\frac{1}{x})+\pi \right)}{4ft}\text{, }&f\neq 0\text{ and }x\neq 0\text{ and }t\neq 0\text{ and }x\neq -1\\d\in \mathrm{R}\text{, }&\left(t=0\text{ or }f=0\right)\text{ and }x=1\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x+1.

$$f\times 1dt=\left(x+1\right)\int _{1}^{\frac{1}{x}}\frac{1}{1+t^{2}}\mathrm{d}t$$

Use the distributive property to multiply x+1 by \int _{1}^{\frac{1}{x}}\frac{1}{1+t^{2}}\mathrm{d}t.

$$f\times 1dt=x\int _{1}^{\frac{1}{x}}\frac{1}{1+t^{2}}\mathrm{d}t+\int _{1}^{\frac{1}{x}}\frac{1}{1+t^{2}}\mathrm{d}t$$

Reorder the terms.

$$dft=x\int _{1}^{\frac{1}{x}}\frac{1}{t^{2}+1}\mathrm{d}t+\int _{1}^{\frac{1}{x}}\frac{1}{t^{2}+1}\mathrm{d}t$$

The equation is in standard form.

$$ftd=\frac{x\left(4\left(\lim_{t\to \frac{1}{x}}\arctan(t)\right)-\pi \right)+4\left(\lim_{t\to \frac{1}{x}}\arctan(t)\right)-\pi }{4}$$

Divide both sides by ft.

$$\frac{ftd}{ft}=\frac{\left\{\begin{matrix}\frac{x\left(4\arctan(\frac{1}{x})-\pi \right)+4\arctan(\frac{1}{x})-\pi }{4},&\\\text{Indeterminate},&\end{matrix}\right.}{ft}$$

Dividing by ft undoes the multiplication by ft.

$$d=\frac{\left\{\begin{matrix}\frac{x\left(4\arctan(\frac{1}{x})-\pi \right)+4\arctan(\frac{1}{x})-\pi }{4},&\\\text{Indeterminate},&\end{matrix}\right.}{ft}$$

Divide \left\{\begin{matrix}\frac{x\left(4\arctan(\frac{1}{x})-\pi \right)+4\arctan(\frac{1}{x})-\pi }{4},&\\\text{Indeterminate},&x=0\end{matrix}\right. by ft.

$$d=\left(\left\{\begin{matrix}\frac{4x\arctan(\frac{1}{x})+4\arctan(\frac{1}{x})-\pi x-\pi }{4ft},&\\\text{Indeterminate},&\end{matrix}\right.\right)$$

Solve for f

$$\left\{\begin{matrix}f=-\frac{\left(x+1\right)\left(-4\arctan(\frac{1}{x})+\pi \right)}{4dt}\text{, }&x\neq -1\text{ and }t\neq 0\text{ and }d\neq 0\text{ and }x\neq 0\\f\in \mathrm{R}\text{, }&\left(d=0\text{ or }t=0\right)\text{ and }x=1\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x+1.

$$f\times 1dt=\left(x+1\right)\int _{1}^{\frac{1}{x}}\frac{1}{1+t^{2}}\mathrm{d}t$$

Use the distributive property to multiply x+1 by \int _{1}^{\frac{1}{x}}\frac{1}{1+t^{2}}\mathrm{d}t.

$$f\times 1dt=x\int _{1}^{\frac{1}{x}}\frac{1}{1+t^{2}}\mathrm{d}t+\int _{1}^{\frac{1}{x}}\frac{1}{1+t^{2}}\mathrm{d}t$$

Reorder the terms.

$$dft=x\int _{1}^{\frac{1}{x}}\frac{1}{t^{2}+1}\mathrm{d}t+\int _{1}^{\frac{1}{x}}\frac{1}{t^{2}+1}\mathrm{d}t$$

The equation is in standard form.

$$dtf=\frac{x\left(4\left(\lim_{t\to \frac{1}{x}}\arctan(t)\right)-\pi \right)+4\left(\lim_{t\to \frac{1}{x}}\arctan(t)\right)-\pi }{4}$$

Divide both sides by dt.

$$\frac{dtf}{dt}=\frac{\left\{\begin{matrix}\frac{x\left(4\arctan(\frac{1}{x})-\pi \right)+4\arctan(\frac{1}{x})-\pi }{4},&\\\text{Indeterminate},&\end{matrix}\right.}{dt}$$

Dividing by dt undoes the multiplication by dt.

$$f=\frac{\left\{\begin{matrix}\frac{x\left(4\arctan(\frac{1}{x})-\pi \right)+4\arctan(\frac{1}{x})-\pi }{4},&\\\text{Indeterminate},&\end{matrix}\right.}{dt}$$

Divide \left\{\begin{matrix}\frac{x\left(4\arctan(\frac{1}{x})-\pi \right)+4\arctan(\frac{1}{x})-\pi }{4},&\\\text{Indeterminate},&x=0\end{matrix}\right. by dt.

$$f=\left(\left\{\begin{matrix}\frac{4x\arctan(\frac{1}{x})+4\arctan(\frac{1}{x})-\pi x-\pi }{4dt},&\\\text{Indeterminate},&\end{matrix}\right.\right)$$