Explore topic-wise InterviewSolutions in .

Assume,

logx = t 

d(logx) = dt 

\(\frac{1}{x}\)dx = dt 

Put t and dt in given equation we get,

⇒ ∫\(\frac{dt}{t}\)

= ln|t| + c. 

But,

t = logx 

= ln| logx| + c

Assume,

e^x + x = t 

d(e^x + x) = dt 

e^x + 1 = dt 

Put t and dt in given equation we get,

⇒ ∫\(\frac{dt}{t}\)

= ln|t| + c. 

But, 

t = e^x + x 

= ln| e^x + 1| + c

Assume,

3 + logx = t 

d(3 + logx) = dt 

⇒ \(\frac{1}{x}\)dx = dt

Put t and dt in given equation we get,

⇒ ∫\(\frac{dt}{t}\)

= ln|t| + c. 

But,

t = 3 + logx 

= ln|3 + logx| + c

∫ sec⁶x dx

In this integral we will use the formula,

1+tan²x = sec²x

I = ∫sec²x sec⁴x dx 

= ∫sec²x (1 + tan²x)²dx 

Now, 

Put tan x = t which means sec²xdx = dt, 

I = ∫(1+ t²)²dt 

= ∫(1+t⁴+2t²) dt 

Now put the value of t, 

which is t = tan x in above integral-

I = tanx + \(\frac{tan^5x}{5}\)+ 2.\(\frac{tan^3x}{3}\)

∫ (x+1)² e^x dx

y = ∫(x² + 2x + 1) e^x dx 

y = ∫(x² + 2x)e^x dx + ∫e^x dx 

We know that,

∫(f(x) +f’(x))e^x dx = f(x) e^x 

Here, 

f(x) = x² 

Then,

f’(x) = 2x 

y = x²e^x + e^x + c 

y = (x² + 1)e^x + c

∫ log(logx)/x dx

Let, 

log x = t 

Differentiating both side with respect to t,

\(\frac{1}{x}\frac{dx}{dt}\) = 1

⇒ \(\frac{dx}{x}\) = dt

Note :-  Always use direct formula for ∫log x dx 

y = ∫log t dt 

y = t log t – t + c 

Again, 

Put t = log x 

y = (log x)log(log x) – log x + c

∫ tan⁵xsec³xdx

In this integral we will use the formula,

1+ tan²x = sec²x 

Then equation will be transform in below form-

I = ∫tan⁵x sec²x secx dx 

= ∫secx tan⁵x sec²x dx 

Now,

Put tan x = t which means sec²xdx = dt,

I = \(\int t^5.\sqrt{1+t^2}\) dt

In this above integral, 

Put 1+t² = k² that is mean tdt = kdk 

I = ∫(k⁴ + 1- 2k) k² dk 

= ∫ (k⁶ + k² – 2k³)dk

= \(\frac{k^7}{7}\)+ \(\frac{k^3}{3}\)- \(\frac{k^4}{2}\)

Now, 

Put the value of k = (1+t²) = sec²x in above equation-

I = \(\frac{sec^{14}x}{7}\)+ \(\frac{sec^{6}x}{3}\)-\(\frac{sec^{8}x}{2}\)

Option : (C)

Given,

∫ (e^x(1+x))/(cos²(xe^x))dx

Let (t) = xe^x

\(\frac{dt}{dx}\) = e^x(1 + x)

⇒ \(\int\frac{dt}{(cos\,t)^2}\) = \(\int\) (sec t)² dt

= tan t 

(put (t) = x ) 

= tan (xe^x) + c

We know that,

\(\int\sqrt{a^2-x^2}dx\) = \(\frac{x}2\)\(\sqrt{a^2-x^2}\) + \(\frac{a^2}2\)sin^-1\(\frac{x}a\) + c

 \(\int\sqrt{2^2-x^2}dx\) = \(\frac{x}2\)\(\sqrt{4-x^2}\) + 2sin^-1\(\frac{x}2\) + c

To find: \(\int\cfrac{dx}{(e^x+e^{-x})}\)

Formula Used:\(\int\cfrac{dx}{1+x^2}=tan^{-1}x\)

Given equation is:

\(\int\cfrac{dx}{(e^x+e^{-x})}=\int\cfrac{e^x dx}{(e^2x+1)}\)....(1)

Let y = e^x … (1)

Differentiating both sides, we get

dy = e^x dx

Substituting in (1),

\(\int\cfrac{dy}{y^2+1}\)

⇒ tan^-1 y

From (1),

⇒ tan^-1 (e^x)

Therefore,

\(\int\cfrac{dx}{(e^x+e^{-x})}=tan^{-1}(e^x)+c\)

Correct answer is (a) \(e^{sin^{-1}}x . (sin^{-1}x - 1) + c\)

To find: \(\int\cfrac{sinx}{(1+cos^2x)}dx\)

Formula Used: \(\int\cfrac{dx}{a^2+x^2}=\cfrac{1}{a}tan^{-1}(\cfrac{x}{a})+c\)

Let y = cos x … (1)

Differentiating both sides, we get 

dy = –sin x dx 

Substituting in given equation,

\(\Rightarrow\)\(\int\cfrac{-dy}{1+y^2}\)

⇒ – tan -1 y 

From (1), 

⇒ –tan -1 (cos x) 

Therefore,

\(\int\cfrac{sinx}{1+cos^2x}dx=-tan^{-1}{(cosx)}+C\)

Suppose sin x = t

d(sin x) = dt = cos x dx

Putting t = sin x and dt = cos x dx in given equation

∫sin⁵x cos x dx = ∫t⁵ dt

By integrating

= t⁶/6 + c

On substituting the value of t

= sin⁶ x/6 + c

I = \(\int\begin{bmatrix}\frac{(x-2)-2}{(x-2)^3} \\[0.3em]\end{bmatrix}e^xdx\)

\(= \int\frac{e^x}{(x-2)^2}dx-\)\(2\int\frac{e^x}{(x-2)^3}dx\)

\(=\frac{e^x}{(x-2)^2}+\)\(2\int\frac{e^x\,dx}{(x-2)^3}-\)\(2\int\frac{e^x\,dx}{(x-2)^3}\)

\(=\frac{e^x}{(x-2)^2}+\) C

Let I = ∫(e^x + 1)² e^x dx

Let e^x +1= t = e^x dx = dt

I = ∫ e^x + 1)² e^x dx 

= ∫t² dt

= \(\frac{t^3}{3}\) 

Now, 

Substitute the value of t.

Hence,

I = \(\frac{(e^x+1)^3}{3}\) + C

∫x² - \(\frac{6}{x}\) + 5e^xdx

= \(\frac{x^3}{x}\) - 6logx + 5e^x + C

∫7^{6log ₇x}dx

= ∫7^{log ₇x^6}dx

= ∫x⁶dx = \(\frac{x^7}{7} + c\)

∫cot²x dx = ∫(cosec²x – 1)dx 

= cot x – x + c

∫7x² – 4sec²x dx = 7.\(\frac{x^3}{3}\) – 4.tanx + c

Suppose I = ∫tan³x sec² x dx

Suppose tan x = t, now

sec² x dx = dt

On substituting the value of x 

I = ∫t³ dt

On integrating 

I = t⁴/4 + c

On substituting the value of t

I = tan⁴x/4 + c

Assume,

logx = t 

d(logx) = dt 

⇒ \(\frac{1}{x}\) dx = dt

Substituting t and dt 

⇒ \(\int\)sint dt

= - cost + c 

But,

t = logx 

⇒ cos(logx) + c.

Assume,

tan^-1x = t 

d( tan^-1x) = dt

⇒ \(\frac{1}{x^2+1}\) = dt

Substituting t and dt 

⇒ \(\int\) e^mt dt

⇒ \(\frac{e^{mt}}{m}\)+ c

But, 

t = tan^-1x

⇒ \(\frac{e^{mtan^{-1}x}}{m}\) + c.

Assume, 

sin^-1x = t 

d( sin^-1x) = dt 

⇒ \(\frac{dx}{\sqrt{1-x^2}}\) = dt

∴ Substituting t and dt in given equation we get 

⇒ \(\int t^3\) dt

⇒  \(\frac{t^4}{4}\) + c

But,

t = sin^-1x

⇒ \(\frac{(sin^{-1}x)^4}{4}\) + c

Assume,

tan^-1x² = t 

d( tan^-1x²) = dt

⇒ \(\frac{2x}{x^4+1}\) = dt

⇒ \(\frac{x}{x^4+1}\) = \(\frac{dt}{2}\)

Substituting t and dt 

⇒ \(\frac{1}{2}\)\(\int\)t dt

⇒ \(\frac{t^2}{4}\) + c

But,

t = tan^-1x² 

⇒ \(\frac{(tan^{-1}x^2)^2}{4}\) + c.

Multiplying and dividing the numerator by e we get the given as

⇒ \(\frac{1}{e}\)\(\int\frac{e^{x-1}+x^{e-1}}{e^x+x^6}\)dx  …(1)

Assume,

e^x + x^e = t 

⇒ d(e^x + x^e )= dt 

⇒ e^x + ex^e-1 = dt 

Substituting t and dt in equation 1 we get 

⇒ \(\frac{1}{e}\int\frac{dt}{t}\)

= ln|t| + c 

But,

t = e^x + x^e 

∴ ln| e^x + x^e | + c.

Let I = ∫ 2x³\(e^{x^2}\)dx

Put x² = t 

2xdx = dt

I = \(\int t\,e^tdt\)

Using integration by parts,

= \(t\int e^tdt\) - \(\int\frac{d}{dt}t\int e^t\,dt\)

We have,

\(\int e^xdx\) = e^x

= te^t - e^t + c

= e^t(t - 1) + c

Substitute value for t,

I = \(e^{x^2}\)(x² - 1) + c

Assume,

logx = t 

⇒ d(logx) = dt 

⇒ \(\frac{1}{x}\)dx = dt

Substituting t and dt in above equation we get,

⇒ \(\int\)t.dt

⇒ \(\frac{t^2}{2}+c\)

But,

t = log(x) 

⇒ \(\frac{log^2x}{2}\) + c  .

Let ,

x = tanθ 

So, 

dx = sec 2θ dθ 

And,

θ  = tan^-x

Putting above values,

\(\int\frac{\sqrt{1+x^2}}{x^4}\)dx 

= \(\int\frac{\sqrt{1+tam^2\theta}}{tan^4\theta}\) sec²θ dθ

= ∫ sec²θ/tan²θ dθ

= ∫ cosec²θ dθ

= - cotθ + c

Put,

θ =  tan^-x

= - cotθ + c

= - cot tan^-x + c

∫cot²(5x + 3)dx = ∫cosec²(5x + 3) -1)dx = \(\frac{cot(5x + 3)}{5}\) - x + c.

∫x cosec²x dx.

= x(-cotx) – ∫-cotx . dx 

= -x cotx + logsinx + C

∫x sinx dx

= x . (-cosx) – ∫-cosx .dx

= -x cosx + sinx + c

∫x²cosxdx = x² – ∫sinx(2x)dx 

= x² -2∫xsinxdx 

= x² - 2[x(-cos x) – ∫-cosxdx] 

= x²sinx + cosx – 2 sinx + c

∫cosec(3 – 5x)cot(3 – 5x)dx

= \(-\frac{cos(3 - 5x)}{-5} + c\)

∫sec²(x – 5)dx = tan(x – 5) + c

∫3^{5x – 3}dx

= \(\frac{3^{5x - 3}}{5.log3} + c\)

∫e^{3 – 4x}dx

= \(\frac{e^{3 - 4x}}{-4} + c\)

∫\(\frac{cosx}{1 + cosx}\)dx

= ∫(cot x . cosecx – cot²x)dx 

= -cosec x – ∫(cosec²x – 1)dx 

= -cosecx -(- cotx) + x + c 

= -cosecx + cotx + x + c

∫cosx.\(\sqrt{cos^2x - 1}\)dx

= ∫cose^x.\(\sqrt{cot^2x}\) dx= ∫cosecx. cot x .dx 

= cosec x + c

Let (x + cos² x) = t 

(1 - sin 2x) dx = dt

\(\int\frac{dt}t\) = Int + c

\(\int\frac{(1-sin2x)}{(x+cos^2x)}dx\) = In (|x + cos²x|) + c

∫\(\sqrt{1 + sin2x}\)dx

= \(\int\sqrt{(cosx + sinx)^2}\)dx

=  ∫(cosx + sinx)dx

= sinx – cosx + c

\(\int\frac{sin2x}{1 + cos^2x}dx\)

= \(\int\frac{-dt}{t}\)

= -logt + c 

= -log(1 + cos²x) + c 

put 1 + cos²x = t² 

cosx(-sinx) dx = dt 

-sin2x dx = dt 

sin2xdx = -dt

\(\int\frac{cosx}{2 + sinx}dx\) = log(2 + sin x) + c

\(\int\frac{cotx}{3 + log(sinx)}dx\)

= \(\frac{dt}{t}\)

= log t + c = log(3 + log(sin x)) + c 

put 3 + log(sin x) = t

\(\frac{cosx}{sinx}dx\) = dt

cos x dx = dt

\(\int\frac{cos^2x.cotx}{4 + 5cos^2x}dx\)

= \(\int\frac{\frac{-1}{10}dx}{t} = \frac{-1}{10}\)logt + c

= log(4 + 5cosec²x) + c 

put 4 + 5 cosec²x = t 

– 10 cosec x – cosec x × cot x dx = dt 

cosec²x . cotxdx = \(\frac{-1}{10}\)dt

\(\int\frac{9x^8 + 9^xlog9}{x^9 + 9^x}dx\) = log(x⁹ + 9^x) + c

\(\int\frac{1}{x(x + 1)(x + 2)}dx\) = \(\frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x + 2}.....(1)\)

1 = A ( x + 1) (x + 2) + B(x) (x + 2) + c × x(x + 1) 

put x = 0, 1 = A(1)(2) ⇒ 1 = 2A ⇒ A = \(\frac{1}{2}\)

put x = -1, 1 = A(0) + B(-1) (-1 + 2) + c(0) ⇒ 1 = -B ⇒ B = -1 

put x = -2, 1 = A(0) + B(0) + c(-2), (-2 + 1) ⇒ 1 = 2c ⇒ c = \(\frac{1}{2}\)

∴ \(\int\frac{dx}{x(x + 1)(x + 2)} = \int\frac{\frac{1}{2}}{x} + \frac{-1}{x + 1} + \frac{\frac{1}{2}}{x + 2}dx\)

= \(\frac{1}{2}\)log2 – log(x + 1) + \(\frac{1}{2}\)log(x + 2) +c

\(\int\frac{3x^2}{1 + x^3}dx\) = log(1 + x³)+c or using

\(\int\frac{f'(x)}{f(x)}dx\) logf(x) + c

put 1 + x³ = t

∴ 3x²dx = dt 

∴ = ∫ \(\frac{1}{t}\)dt = log t + c = log(1 + x³) + c

\(\int\frac{e^x - 1}{e^x - x}dx\) = log(e^x – x) + c

\(\int\frac{4x + 3}{2x^2 + 3x + 5}dx\)

= log(2x² + 3x +5) + c 

∴ \(\int\frac{f'(x)}{f(x)}dx\) = log(f(x)) + C

∫sec x (sec x – tan x)dx 

= ∫sec²x – sec x .tan x. dx 

= tan x – sec x + c