# Get Answers to all your Questions

### Answers (2)

Answer: $\sqrt{x^{2}+1}-\log \left|x+\sqrt{x^{2}+1}\right|+c$

Given: $\int \frac{x-1}{\sqrt{x^{2}+1}} d x$

Hint: Simplify the given function

Solution:

\begin{aligned} &I=\int \frac{x-1}{\sqrt{x^{2}+1}} d x \\ &I=\frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}+1}} d x-\int \frac{1}{\sqrt{x^{2}+1}} d x \\ &I=\frac{1}{2}\left[\frac{\sqrt{x^{2}+1}}{\frac{1}{2}}\right]-\log \left|x+\sqrt{x^{2}+1}\right|+c \end{aligned}

$\left[\begin{array}{l} U \sin g \\ \int(f(x))^{n} f^{1}(x) d x=\frac{[f(x)]^{n+1}}{n+1}+c \\ \int \frac{1}{\sqrt{x^{2}+a^{2}}} d x=\log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \end{array}\right]$

$I=\sqrt{x^{2}+1}-\log \left|x+\sqrt{x^{2}+1}\right|+c$

View full answer

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support

Answer: $\sqrt{x^{2}+1}-\log \left|x+\sqrt{x^{2}+1}\right|+c$

Given: $\int \frac{x-1}{\sqrt{x^{2}+1}} d x$

Hint: Simplify the given function

Solution:

\begin{aligned} &I=\int \frac{x-1}{\sqrt{x^{2}+1}} d x \\ &I=\frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}+1}} d x-\int \frac{1}{\sqrt{x^{2}+1}} d x \\ &I=\frac{1}{2}\left[\frac{\sqrt{x^{2}+1}}{\frac{1}{2}}\right]-\log \left|x+\sqrt{x^{2}+1}\right|+c \end{aligned}

\begin{aligned} &{\left[\begin{array}{l} U \sin g \\ \int(f(x))^{n} f^{1}(x) d x=\frac{[f(x)]^{n+1}}{n+1}+c \\ \int \frac{1}{\sqrt{x^{2}+a^{2}}} d x=\log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \end{array}\right]} \\ &I=\sqrt{x^{2}+1}-\log \left|x+\sqrt{x^{2}+1}\right|+c \end{aligned}

View full answer