Name: Standard Indefinite Integrals
Uploaded: 2019-07-19
Duration: 3 min 29 s
Description: How does the method of substitution help in finding some standard trigonometric integration? Here is the answer. Watch the whole video to understand the conc...

Question

Solve

\int \frac{x}{\sqrt{x^{2} + x + 1}} d x

.

Accepted Answer

Let $I = \int \frac{x}{\sqrt{x^{2} + x + 1}} d x$

Let $x \equiv A \frac{d}{d x} (x^{2} + x + 1) + B$

$\Rightarrow x \equiv A (2 x + 1) + B$

Equating coefficients of similar powers of $x$ , we get-

$1 = 2 A \Rightarrow A = \frac{1}{2}$

$A + B = 0 \Rightarrow B = - A \Rightarrow B = - \frac{1}{2}$

$∴ I = \int \frac{2 x + 1}{2 \sqrt{x^{2} + x + 1}} d x - \int \frac{d x}{2 \sqrt{x^{2} + x + 1}}$

$= I_{1} - I_{2}$

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

Let $x^{2} + x + 1 = t \Rightarrow (2 x + 1) d x = d t$

$∴ I_{1} = \frac{1}{2} \int \frac{d t}{\sqrt{t}} = \frac{1}{2} \cdot 2 \sqrt{t} = \sqrt{t} + c_{1}$

$= \sqrt{x^{2} + x + 1} + c_{1}$

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

$= \frac{1}{2} \log |x + \frac{1}{2} + \sqrt{x^{2} + x + 1}| + c_{2}$

$∴ I = I_{1} - I_{2}$

$= \sqrt{x^{2} + x + 1} - \frac{1}{2} \log |x + \frac{1}{2} + \sqrt{x^{2} + x + 1}| + c$

Solve $\int \frac{x}{\sqrt{x^{2} + x + 1}} d x$ .

Important Questions on Indefinite Integrals - II

Learn and Prepare for any exam you want !

Solve ∫xx2+x+1dx.

Important Questions on Indefinite Integrals - II

Learn and Prepare for any exam you want !

Solve $\int \frac{x}{\sqrt{x^{2} + x + 1}} d x$ .