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

\int \frac{d x}{(x + 1)^{2} (x^{2} + 1)}

Accepted Answer

We have to find $\int \frac{d x}{{(x + 1)}^{2} (x^{2} + 1)} .$

$\Rightarrow \frac{1}{{(x + 1)}^{2} (x^{2} + 1)} = \frac{A x + B}{x^{2} + 1} + \frac{C}{x + 1} + \frac{D}{{(x + 1)}^{2}} \Rightarrow 1 = (A x + B) {(1 + x)}^{2} + C (x^{2} + 1) (x + 1) + D (x^{2} + 1) \Rightarrow 1 = (A x + B) (x^{2} + 2 x + 1) + C (x^{3} + x^{2} + x + 1) + D (x^{2} + 1) \Rightarrow 1 = (A x^{3} + (2 A + B) x^{2} + (A + 2 B) x + B) + C (x^{3} + x^{2} + x + 1) + D (x^{2} + 1)$

Equating the coefficients of $x^{3}, x^{2}, x$ and constant terms, we get

$\Rightarrow A + C = 0, 2 A + B + C + D = 0, A + 2 B + C = 0$ and $C + B + D = 1$

$\Rightarrow A = - \frac{1}{2}, B = 0, C = \frac{1}{2}, D = \frac{1}{2}$

$∴ \frac{1}{{(1 + x)}^{2} (1 + x^{2})} = \frac{- \frac{x}{2}}{x^{2} + 1} + \frac{1}{2} (\frac{1}{1 + x}) + \frac{1}{2} (\frac{1}{{(x + 1)}^{2}})$

$\Rightarrow \frac{1}{{(1 + x)}^{2} (1 + x^{2})} = - \frac{1}{2} (\frac{x}{x^{2} + 1}) + \frac{1}{2} (\frac{1}{x + 1}) + \frac{1}{2} (\frac{1}{{(x + 1)}^{2}})$

Integrating with respect to $x$ ,

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

$\Rightarrow \int \frac{d x}{{(x + 1)}^{2} (x^{2} + 1)} = I_{1} + \frac{1}{2} \int \frac{d x}{x + 1} + I_{3}$

Solving $I_{1}$

$I_{1} = - \frac{1}{2} \int \frac{x}{x^{2} + 1} d x$

Put $x^{2} + 1 = t$

Differentiating both sides with respect to $x$ ,

$2 x + 0 = \frac{d t}{d x} \Rightarrow \frac{d t}{2 x} = d x$

Hence, we can write

$I_{1} = - \frac{1}{2} \int \frac{x}{t} \frac{d t}{2 x} = - \frac{1}{2 \times 2} \int \frac{d t}{t} = - \frac{1}{4} \log |t| + c_{1} \Rightarrow I_{1} = - \frac{1}{4} \log |x^{2} + 1| + c_{1}$

Solving $I_{3}$

$I_{3} = \frac{1}{2} \int \frac{d x}{{(x + 1)}^{2}} = - \frac{1}{2} \cdot \frac{1}{x + 1} + c_{2} (∵ \int x^{n} = \frac{x^{n + 1}}{n + 1} + c, where n \neq - 1)$

Hence

$\int \frac{d x}{{(x + 1)}^{2} (x^{2} + 1)} = I_{1} + \frac{1}{2} \int \frac{d x}{x + 1} + I_{3}$

$\Rightarrow \int \frac{1}{{(1 + x)}^{2} (1 + x^{2})} d x = - \frac{1}{4} \log |x^{2} + 1| + c_{1} + \frac{1}{2} \log (x + 1) - \frac{1}{2} \cdot \frac{1}{x + 1} + c_{2} \Rightarrow \int \frac{1}{{(1 + x)}^{2} (1 + x^{2})} d x = \frac{1}{2} \log (x + 1) - \frac{1}{2 (x + 1)} - \frac{1}{4} \log |x^{2} + 1| + c$

Evaluate $\int \frac{d x}{(x + 1)^{2} (x^{2} + 1)}$

Important Questions on Integrals

Learn and Prepare for any exam you want !

Evaluate ∫dx(x+1)2 (x2+1)

Important Questions on Integrals

Learn and Prepare for any exam you want !

Evaluate $\int \frac{d x}{(x + 1)^{2} (x^{2} + 1)}$