Name: Second Fundamental Theorem of Calculus
Uploaded: 2019-07-19
Duration: 5 min 22 s
Description: In this video, we will learn about the evaluating the definite integral by using the second fundamnetal theorem of calculus in details.

Question

Evaluate the integral

\int x \tan^{- 1} (2 x + 3) d x

.

Accepted Answer

Let $I = \int x \tan^{- 1} (2 x + 3) d x$

$= \tan^{- 1} (2 x + 3) \frac{x^{2}}{2} - \int \frac{2}{1 + {(2 x + 3)}^{2}} . \frac{x^{2}}{2} d x + C$

$= \frac{x^{2}}{2} \tan^{- 1} (2 x + 3) - \int \frac{x^{2}}{4 x^{2} + 12 x + 10} d x + C$

$\frac{x^{2}}{2} \tan^{- 1} (2 x + 3) - \int [\frac{1}{4} - \frac{3 x + \frac{5}{2}}{4 x^{2} + 12 x + 10}] d x$

$= \frac{x^{2}}{2} \tan^{- 1} (2 x + 3) - \frac{x}{4} + \int \frac{(3 x + \frac{5}{2})}{2 (2 x^{2} + 6 x 5)} d x + C$

$= \frac{x^{2}}{2} \tan^{- 1} (2 x + 3) - \frac{x}{4} + \frac{3}{2} \int \frac{(x + \frac{5}{6}) d x}{2 x^{2} + 6 x + 5} + C$

$= \frac{x^{2}}{2} \tan^{- 1} (2 x + 3) - \frac{x}{4} + \frac{3}{8} \int \frac{(4 x + \frac{10}{3}) d x}{2 x^{2} + 6 x + 5} + C$

$= \frac{x^{2}}{2} \tan^{- 1} (2 x + 3) - \frac{x}{4} + \frac{3}{8} \int \frac{(4 x + 6 - \frac{8}{3}) d x}{2 x^{2} + 6 x + 5} + C$

$= \frac{x^{2}}{2} \tan^{- 1} (2 x + 3) - \frac{x}{4} + \frac{3}{8} \int \frac{(4 x + 6) d x}{2 x^{2} + 6 x + 5} - \int \frac{d x}{2 x^{2} + 6 x + 5} + C$

$= \frac{x^{2}}{2} \tan^{- 1} (2 x + 3) - \frac{x}{4} + \frac{3}{8} \log | 2 x^{2} + 6 x + 5 | - \frac{1}{2} \int \frac{d x}{x^{2} + 3 x + \frac{5}{2}} + C$

$[\int \frac{\int_{}^{'} (x)}{\int (x)} d x = \log | \int (x) | + C]$

$= \frac{x^{2}}{2} \tan^{- 1} (2 x + 3) - \frac{x}{4} + \frac{3}{8} \log | 2 x^{2} + 6 x + 5 | - \frac{1}{2} \int \frac{d x}{x 2 + 3 x + \frac{9}{4} - \frac{9}{4} + \frac{5}{2}}$

$= \frac{x^{2}}{2} \tan - 1 (2 x + 3) - \frac{x}{4} + \frac{3}{8} \log | 2 x^{2} + 6 x + 5 | - \frac{1}{2} \int \frac{d x}{{(x + \frac{3}{2})}^{2} + {(\frac{1}{2})}^{2}}$

$= \frac{x^{2}}{2} \tan - 1 (2 x + 3) - \frac{x}{4} + \frac{3}{8} | 2 x^{2} + 6 x + 5 | - \frac{1}{2} \times \frac{1}{\frac{1}{2}} \tan^{- 1} (2 x + 3) + C$

$= \frac{1}{2} (x^{2} - 2) \tan - 1 (2 x + 3) - \frac{x}{4} + \frac{3}{8} \log | 2 x^{2} + 6 x + 5 | + C$

Evaluate the integral $\int x \tan^{- 1} (2 x + 3) d x$ .

Important Questions on Integrals

Learn and Prepare for any exam you want !

Evaluate the integral ∫ xtan-1(2x+3)dx.

Important Questions on Integrals

Learn and Prepare for any exam you want !

Evaluate the integral $\int x \tan^{- 1} (2 x + 3) d x$ .