First slide

Introduction to integration

Question

$\int \tan^{- 1} \sqrt{x} dx$ is equal to

Moderate

A

${xtan}^{- 1} \sqrt{x} - \sqrt{x} + \frac{1}{2} \tan^{- 1} \sqrt{x} + C$
B

${xtan}^{- 1} \sqrt{x} + C$
C

$- \sqrt{x} + {xtan}^{- 1} \sqrt{x} + C$
D

None of these above $I_{1} = \int \frac{\sqrt{x}}{x + 1} dx$

Solution

$\begin{array}{l} I = \int 1 \cdot \tan^{- 1} \sqrt{x} dx \\ = \tan^{- 1} \sqrt{x} \cdot x - \int \frac{1}{1 + x} \cdot \frac{1}{2 \sqrt{x}} \cdot xdx \\ = {xtan}^{- 1} \sqrt{x} - \frac{1}{2} \int \frac{\sqrt{x}}{1 + x} dx \end{array}$
Let $I_{1} = \int \frac{\sqrt{x}}{x + 1} dx$
Assume that $\sqrt{x} = t$
$\begin{array}{l} \Rightarrow \frac{1}{2 \sqrt{x}} dx = dt \Rightarrow dx = 2 tdt \\ ∴ I_{1} = \int \frac{t \cdot 2 t}{1 + t^{2}} dt = 2 \int (\frac{1 + t^{2}}{1 + t^{2}} - \frac{1}{1 + t^{2}}) dt \\ = 2 t - 2 \tan^{- 1} t \\ = 2 \sqrt{x} - 2 \tan^{- 1} \sqrt{x} \\ ∴ I = {xtan}^{- 1} \sqrt{x} - \frac{1}{2} (2 \sqrt{x} - 2 \tan^{- 1} \sqrt{x}) + C \\ = {xtan}^{- 1} \sqrt{x} - \sqrt{x} + (\tan^{- 1} \sqrt{x}) + C \\ = (x + 1) \tan^{- 1} \sqrt{x} - \sqrt{x} + C \end{array}$

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