If \int \frac1\sqrt[5](x-1)^4(x+3)^6 ~d x=A(\frac\alpha x-1\beta x+3)^B+C , where C is the constant of integration, then the value of \alpha+\beta+20 AB is \underline\hspace2cm .

Indefinite Integrals PYQ — JEE Main 2024 · 8 Apr · Shift 2 · Q53 with Solution

View written solutionFree

Correct answer: 7

Rewrite the integrand

Given

I=\int \frac{1}{\sqrt[5]{(x-1)^4(x+3)^6}}\,dx

we write it as

I=\int (x-1)^{-4/5}(x+3)^{-6/5}\,dx.

Look for a substitution suggested by the final form

Since the answer is of the form

A\left(\frac{\alpha x-1}{\beta x+3}\right)^B + C,

it is natural to try

t=\frac{x-1}{x+3}.

Then

\frac{dt}{dx}=\frac{(x+3)-(x-1)}{(x+3)^2}=\frac{4}{(x+3)^2},

dx=\frac{(x+3)^2}{4}\,dt.

Also,

x-1=t(x+3).

Hence the integrand becomes

(x-1)^{-4/5}(x+3)^{-6/5} =\bigl(t(x+3)\bigr)^{-4/5}(x+3)^{-6/5} =t^{-4/5}(x+3)^{-10/5} =t^{-4/5}(x+3)^{-2}.

Therefore,

I=\int t^{-4/5}(x+3)^{-2}\cdot \frac{(x+3)^2}{4}\,dt =\frac14\int t^{-4/5}\,dt.

Integrate

\frac14\int t^{-4/5}\,dt =\frac14\cdot \frac{t^{1/5}}{1/5}+C =\frac54 t^{1/5}+C.

Substituting back,

I=\frac54\left(\frac{x-1}{x+3}\right)^{1/5}+C.

Match with the required form

We compare

\frac54\left(\frac{x-1}{x+3}\right)^{1/5}+C

with

A\left(\frac{\alpha x-1}{\beta x+3}\right)^B+C.

Thus,

A=\frac54,\quad \alpha=1,\quad \beta=1,\quad B=\frac15.

Compute the required value

\alpha+\beta+20AB=1+1+20\cdot \frac54\cdot \frac15.

Now,

20\cdot \frac54\cdot \frac15 = 5.

So,

\alpha+\beta+20AB=1+1+5=7.

Compare with stored answer

Derived answer = $7$ .

Stored correct answer = $7$ .

They agree.

More from Indefinite Integrals

Let $\int \frac{2 - t a n x}{3 + t a n x} d x = \frac{1}{2} (α x + lo g_{e} ∣ β sin x + γ cos x ∣) + C$ , where $C$ is the constant of integration. Then $α + \frac{γ}{β}$ is equal to :2024 · MCQ
$The integral \int \frac{( x ^{8} - x ^{2} ) d x}{( x ^{12} + 3 x ^{6} + 1 ) t a n ^{- 1} ( x ^{3} + \frac{1}{x ^{3}} )} is equal to :$ 2024 · MCQ
For $x \in (- \frac{π}{2}, \frac{π}{2})$ , if $y (x) = \int \frac{cosec x + s i n x}{cosec x s e c x + t a n x s i n ^{2} x} d x$ , and $lim_{x \to (\frac{π}{2})^{-}} y (x) = 0$ then $y (\frac{π}{4})$ …2024 · MCQ
If $\int \frac{s i n ^{\frac{3}{2}} x + c o s ^{\frac{3}{2}} x}{s i n ^{3} x c o s ^{3} x s i n ( x - θ )} d x = A cos θ tan x - sin θ + B cos θ - sin θ cot x + C$ , where $C$ is the integration constant, then $A B$ …2024 · MCQ
Let $I (x) = \int \frac{x ^{2} ( x s e c ^{2} x + t a n x )}{( x t a n x + 1 ) ^{2}} d x$ . If $I (0) = 0$ , then $I (\frac{π}{4})$ is equal to :2023 · MCQ
Let $I (x) = \int \frac{( x + 1 )}{x ( 1 + x e ^{x} ) ^{2}} d x, x > 0$ . If $lim_{x \to \infty} I (x) = 0$ , then $I (1)$ is equal to :2023 · MCQ
The integral $\int [(\frac{x}{2})^{x} + (\frac{2}{x})^{x}] ln (\frac{e x}{2}) d x$ is equal to :2023 · MCQ
If $I (x) = \int e^{s i n^{2} x} (cos x sin 2 x - sin x) d x$ and $I (0) = 1$ , then $I (\frac{π}{3})$ is equal to :2023 · MCQ

View written solutionFree

Correct answer: 7

Rewrite the integrand

Given

I=\int \frac{1}{\sqrt[5]{(x-1)^4(x+3)^6}}\,dx

we write it as

I=\int (x-1)^{-4/5}(x+3)^{-6/5}\,dx.

Look for a substitution suggested by the final form

Since the answer is of the form

A\left(\frac{\alpha x-1}{\beta x+3}\right)^B + C,

it is natural to try

t=\frac{x-1}{x+3}.

Then

\frac{dt}{dx}=\frac{(x+3)-(x-1)}{(x+3)^2}=\frac{4}{(x+3)^2},

dx=\frac{(x+3)^2}{4}\,dt.

Also,

x-1=t(x+3).

Hence the integrand becomes

(x-1)^{-4/5}(x+3)^{-6/5} =\bigl(t(x+3)\bigr)^{-4/5}(x+3)^{-6/5} =t^{-4/5}(x+3)^{-10/5} =t^{-4/5}(x+3)^{-2}.

Therefore,

I=\int t^{-4/5}(x+3)^{-2}\cdot \frac{(x+3)^2}{4}\,dt =\frac14\int t^{-4/5}\,dt.

Integrate

\frac14\int t^{-4/5}\,dt =\frac14\cdot \frac{t^{1/5}}{1/5}+C =\frac54 t^{1/5}+C.

Substituting back,

I=\frac54\left(\frac{x-1}{x+3}\right)^{1/5}+C.

Match with the required form

We compare

\frac54\left(\frac{x-1}{x+3}\right)^{1/5}+C

with

A\left(\frac{\alpha x-1}{\beta x+3}\right)^B+C.

Thus,

A=\frac54,\quad \alpha=1,\quad \beta=1,\quad B=\frac15.

Compute the required value

\alpha+\beta+20AB=1+1+20\cdot \frac54\cdot \frac15.

Now,

20\cdot \frac54\cdot \frac15 = 5.

So,

\alpha+\beta+20AB=1+1+5=7.

Compare with stored answer

Derived answer = $7$ .

Stored correct answer = $7$ .

They agree.

Indefinite Integrals question

Indefinite Integrals question

More from Indefinite Integrals

Indefinite Integrals question

Indefinite Integrals question

More from Indefinite Integrals