Find the integral of 1-x-3x-4 with respect to X-

By using partial fraction method, ∫ I dx=∫1(x 3)(x+4)dx 1(x 3)(x+4)=Ax 3+Bx+4 nbsp; 1= A(x+4)+B(x 3) On comparing the coefficients of constant and x term, ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Property 1

Find the inte...

Question

Find the integral of $\frac{1}{(x - 3) (x + 4)}$ with respect to $x .$

Open in App

Solution

By using partial fraction method,
$\int I d x = \int \frac{1}{(x - 3) (x + 4)} d x$

$\frac{1}{(x - 3) (x + 4)} = \frac{A}{x - 3} + \frac{B}{x + 4} 1 = A (x + 4) + B (x - 3)$

On comparing the coefficients of constant and $x$ term, we get
$0 = A + B & 4 A - 3 B = 1 \Rightarrow A = \frac{1}{7} & B = \frac{- 1}{7}$

$\Rightarrow \int I d x = \frac{1}{7} \int \frac{1}{x - 3} d x - \frac{1}{7} \int \frac{1}{x + 4} d x$
$= \frac{1}{7} {log}_{e} (x - 3) - \frac{1}{7} {log}_{3} (x + 4)$
$= \frac{1}{7} {log}_{e} (\frac{x - 3}{x + 4}) + C$

Suggest Corrections

Similar questions

Q. Integrate the following function with respect to

x

\frac{(x + 1) . (x + log x)^{4}}{3 x}

Q. Integrate with respect to

x

\frac{3 x + 2}{(x + 1)^{2} (x + 2)}

Q. Integrate with respect to

x

{sec}^{3} x

Q. Find the integral of

\frac{1}{\sqrt{x^{2} + a^{2}}}

with respect to x

Q. Integral of

\frac{1}{\sqrt{x^{2} + 4}}

with respect to

(x^{2} + 3)

is equal to

Join BYJU'S Learning Program

Find the integral of 1(x−3)(x+4) with respect to x.

By using partial fraction method, ∫I dx=∫1(x−3)(x+4)dx 1(x−3)(x+4)=Ax−3+Bx+41=A(x+4)+B(x−3) On comparing the coefficients of constant and x term, we get 0=A+B & 4A−3B=1⇒A=17 & B=−17 ⇒∫I dx=17∫1x−3dx−17∫1x+4dx =17loge(x−3)−17log3(x+4) =17loge(x−3x+4)+C

Find the integral of $\frac{1}{(x - 3) (x + 4)}$ with respect to $x .$