Let I(x)=\int \sqrt\fracx+7x ~d x and I(9)=12+7 \log _e 7 . If I(1)=\alpha+7 \log _e(1+2 \sqrt2) , then \alpha^4 is equal to \underline\hspace2cm .

Indefinite Integrals PYQ — JEE Main 2023 · 12 Apr · Shift 1 · Q43 with Solution

View written solutionFree

Correct answer: 64

Evaluate the indefinite integral

We need

I(x)=\int \sqrt{\frac{x+7}{x}}\,dx=\int \sqrt{1+\frac{7}{x}}\,dx.

A standard substitution works well:

x=7\sinh^2 t.

Then

\frac{x+7}{x}=\frac{7\sinh^2 t+7}{7\sinh^2 t}=\frac{\sinh^2 t+1}{\sinh^2 t}=\frac{\cosh^2 t}{\sinh^2 t}=\coth^2 t,

\sqrt{\frac{x+7}{x}}=\coth t.

Also,

dx=14\sinh t\cosh t\,dt.

Hence

I(x)=\int \coth t\cdot 14\sinh t\cosh t\,dt =14\int \cosh^2 t\,dt.

Using

\cosh^2 t=\frac{1+\cosh 2t}{2},

we get

I(x)=14\int \cosh^2 t\,dt =14\left(\frac{t}{2}+\frac{\sinh 2t}{4}\right)+C =7t+\frac{7}{2}\sinh 2t+C.

Now,

\sinh 2t=2\sinh t\cosh t.

Since $x=7\sinh^2 t$ ,

\sinh t=\sqrt{\frac{x}{7}}, \qquad \cosh t=\sqrt{1+\sinh^2 t}=\sqrt{\frac{x+7}{7}}.

Thus

\frac{7}{2}\sinh 2t=7\sinh t\cosh t =7\sqrt{\frac{x}{7}}\sqrt{\frac{x+7}{7}} =\sqrt{x(x+7)}.

Also,

t=\sinh^{-1}\sqrt{\frac{x}{7}}.

I(x)=\sqrt{x(x+7)}+7\sinh^{-1}\sqrt{\frac{x}{7}}+C.

Using

\sinh^{-1} u=\ln\left(u+\sqrt{u^2+1}\right),

with $u=\sqrt{x/7}$ ,

\sinh^{-1}\sqrt{\frac{x}{7}} =\ln\left(\sqrt{\frac{x}{7}}+\sqrt{\frac{x+7}{7}}\right) =\ln\left(\frac{\sqrt{x}+\sqrt{x+7}}{\sqrt{7}}\right).

Therefore,

I(x)=\sqrt{x(x+7)}+7\ln\left(\frac{\sqrt{x}+\sqrt{x+7}}{\sqrt{7}}\right)+C.

Use the condition $I(9)=12+7\ln 7$ to find $C$

Substitute $x=9$ :

\sqrt{9(16)}=12,

and

\frac{\sqrt{9}+\sqrt{16}}{\sqrt{7}}=\frac{3+4}{\sqrt7}=\frac{7}{\sqrt7}=\sqrt7.

I(9)=12+7\ln(\sqrt7)+C=12+\frac{7}{2}\ln7+C.

Given

I(9)=12+7\ln7,

therefore

\frac{7}{2}\ln7+C=7\ln7 \implies C=\frac{7}{2}\ln7.

Hence

I(x)=\sqrt{x(x+7)}+7\ln\left(\frac{\sqrt{x}+\sqrt{x+7}}{\sqrt{7}}\right)+\frac{7}{2}\ln7.

Combine the logarithms:

I(x)=\sqrt{x(x+7)}+7\ln(\sqrt{x}+\sqrt{x+7}).

Compute $I(1)$

At $x=1$ ,

\sqrt{1(1+7)}=\sqrt8=2\sqrt2.

Also,

\sqrt1+\sqrt8=1+2\sqrt2.

Thus

I(1)=2\sqrt2+7\ln(1+2\sqrt2).

Comparing with

I(1)=\alpha+7\ln(1+2\sqrt2),

we get

\alpha=2\sqrt2.

Therefore,

\alpha^4=(2\sqrt2)^4.

Now,

(2\sqrt2)^2=8 \quad\Rightarrow\quad (2\sqrt2)^4=8^2=64.

Final answer

\boxed{64}

This matches the stored correct answer.

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View written solutionFree

Correct answer: 64

Evaluate the indefinite integral

We need

I(x)=\int \sqrt{\frac{x+7}{x}}\,dx=\int \sqrt{1+\frac{7}{x}}\,dx.

A standard substitution works well:

x=7\sinh^2 t.

Then

\frac{x+7}{x}=\frac{7\sinh^2 t+7}{7\sinh^2 t}=\frac{\sinh^2 t+1}{\sinh^2 t}=\frac{\cosh^2 t}{\sinh^2 t}=\coth^2 t,

\sqrt{\frac{x+7}{x}}=\coth t.

Also,

dx=14\sinh t\cosh t\,dt.

Hence

I(x)=\int \coth t\cdot 14\sinh t\cosh t\,dt =14\int \cosh^2 t\,dt.

Using

\cosh^2 t=\frac{1+\cosh 2t}{2},

we get

I(x)=14\int \cosh^2 t\,dt =14\left(\frac{t}{2}+\frac{\sinh 2t}{4}\right)+C =7t+\frac{7}{2}\sinh 2t+C.

Now,

\sinh 2t=2\sinh t\cosh t.

Since $x=7\sinh^2 t$ ,

\sinh t=\sqrt{\frac{x}{7}}, \qquad \cosh t=\sqrt{1+\sinh^2 t}=\sqrt{\frac{x+7}{7}}.

Thus

\frac{7}{2}\sinh 2t=7\sinh t\cosh t =7\sqrt{\frac{x}{7}}\sqrt{\frac{x+7}{7}} =\sqrt{x(x+7)}.

Also,

t=\sinh^{-1}\sqrt{\frac{x}{7}}.

I(x)=\sqrt{x(x+7)}+7\sinh^{-1}\sqrt{\frac{x}{7}}+C.

Using

\sinh^{-1} u=\ln\left(u+\sqrt{u^2+1}\right),

with $u=\sqrt{x/7}$ ,

\sinh^{-1}\sqrt{\frac{x}{7}} =\ln\left(\sqrt{\frac{x}{7}}+\sqrt{\frac{x+7}{7}}\right) =\ln\left(\frac{\sqrt{x}+\sqrt{x+7}}{\sqrt{7}}\right).

Therefore,

I(x)=\sqrt{x(x+7)}+7\ln\left(\frac{\sqrt{x}+\sqrt{x+7}}{\sqrt{7}}\right)+C.

Use the condition $I(9)=12+7\ln 7$ to find $C$

Substitute $x=9$ :

\sqrt{9(16)}=12,

and

\frac{\sqrt{9}+\sqrt{16}}{\sqrt{7}}=\frac{3+4}{\sqrt7}=\frac{7}{\sqrt7}=\sqrt7.

I(9)=12+7\ln(\sqrt7)+C=12+\frac{7}{2}\ln7+C.

Given

I(9)=12+7\ln7,

therefore

\frac{7}{2}\ln7+C=7\ln7 \implies C=\frac{7}{2}\ln7.

Hence

I(x)=\sqrt{x(x+7)}+7\ln\left(\frac{\sqrt{x}+\sqrt{x+7}}{\sqrt{7}}\right)+\frac{7}{2}\ln7.

Combine the logarithms:

I(x)=\sqrt{x(x+7)}+7\ln(\sqrt{x}+\sqrt{x+7}).

Compute $I(1)$

At $x=1$ ,

\sqrt{1(1+7)}=\sqrt8=2\sqrt2.

Also,

\sqrt1+\sqrt8=1+2\sqrt2.

Thus

I(1)=2\sqrt2+7\ln(1+2\sqrt2).

Comparing with

I(1)=\alpha+7\ln(1+2\sqrt2),

we get

\alpha=2\sqrt2.

Therefore,

\alpha^4=(2\sqrt2)^4.

Now,

(2\sqrt2)^2=8 \quad\Rightarrow\quad (2\sqrt2)^4=8^2=64.

Final answer

\boxed{64}

This matches the stored correct answer.

Indefinite Integrals question

Indefinite Integrals question

More from Indefinite Integrals

Indefinite Integrals question

Indefinite Integrals question

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