Indefinite Integration Questions

MathematicsIndefinite Integration30 Questions Found

MathematicsIntegration by SubstitutionJEE Advanced 1996Moderate

Evaluate $\int \frac{( x + 1 )}{x ( 1 + x e ^{x} ) ^{2}} d x$

Answer:

lo g (\frac{1 + x e ^{x}}{x e ^{x}}) - \frac{1}{1 + x e ^{x}} + C

MathematicsIntegration using Partial FractionsJEE Advanced 1999Moderate

View in:English Hindi

Integrate $\int \frac{x ^{3} + 3 x + 2}{( x ^{2} + 1 ) ^{2} ( x + 1 )} d x$

Answer:

- \frac{1}{2} lo g ∣ x + 1∣ + \frac{1}{4} lo g (x^{2} + 1) + \frac{3}{2} tan^{- 1} x + \frac{x}{1 + x ^{2}} + C

MathematicsIntegration by SubstitutionJEE Advanced 2001Moderate

View in:English Hindi

Evaluate $\int sin^{- 1} (\frac{2 x + 2}{4 x ^{2} + 8 x + 13}) d x$

Answer:

(x + 1) tan^{- 1} (\frac{2 x + 2}{3}) - \frac{3}{4} lo g (4 x^{2} + 8 x + 13) + C

MathematicsIntegration by SubstitutionJEE Advanced 2002Moderate

View in:English Hindi

For any natural number $m$ , evaluate $\int (x^{3 m} + x^{2 m} + x^{m}) (2 x^{2 m} + 3 x^{m} + 6)^{1/ m} d x, x > 0$

Answer:

\frac{1}{6} \frac{( 2 x ^{3 m} + 3 x ^{2 m} + 6 x ^{m} ) ^{\frac{m + 1}{m}}}{m + 1} + C

MathematicsIntegration by SubstitutionJEE Advanced 1995Moderate

View in:English Hindi

The value of the integral $\int \frac{c o s ^{3} x + c o s ^{5} x}{s i n ^{2} x + s i n ^{4} x} d x$ is

(A)

sin x - 6 tan^{- 1} (sin x) + c

(B)

sin x - 2 (sin x)^{- 1} + c

(C)

sin x - 2 (sin x)^{- 1} - 6 tan^{- 1} (sin x) + c

(D)

sin x - 2 (sin x)^{- 1} + 5 tan^{- 1} (sin x) + c

Answer:C

MathematicsIntegration by SubstitutionJEE Advanced 2006Moderate

View in:English Hindi

$\int \frac{x ^{2} - 1}{x ^{3} 2 x ^{4} - 2 x ^{2} + 1} d x =$

(A)

\frac{2 x ^{4} - 2 x ^{2} + 1}{x ^{2}} + c

(B)

\frac{2 x ^{4} - 2 x ^{2} + 1}{x ^{3}} + c

(C)

\frac{2 x ^{4} - 2 x ^{2} + 1}{x} + c

(D)

\frac{2 x ^{4} - 2 x ^{2} + 1}{2 x ^{2}} + c

Answer:D

MathematicsFundamental Integration FormulasJEE Advanced 2007Easy

View in:English Hindi

Let $F (x)$ be an indefinite integral of $sin^{2} x$ . STATEMENT-1: The function $F (x)$ satisfies $F (x + π) = F (x)$ because STATEMENT-2: $sin^{2} (x + π) = sin^{2} x$ for all real $x$ .

(A)

Statement-1 is True, statement-2 is True; Statement-2 is a correct explanation for Statement-1.

(B)

Statement-1 is True, statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.

(C)

Statement-1 is True, Statement-2 is False

(D)

Statement-1 is False, Statement-2 is True.

Answer:D

MathematicsIntegration by SubstitutionJEE Advanced 2007Moderate

View in:English Hindi

Let $f (x) = \frac{x}{( 1 + x ^{n} ) ^{1/ n}}$ for $n \geq 2$ and $g (x) = (f \circ f \circ ... \circ f) (x)$ ( $f$ occurs $n$ times). Then $\int x^{n - 2} g (x) d x$ equals

(A)

\frac{1}{n ( n - 1 )} (1 + n x^{n})^{1 - 1/ n} + K

(B)

\frac{1}{n - 1} (1 + n x^{n})^{1 - 1/ n} + K

(C)

\frac{1}{n ( n + 1 )} (1 + n x^{n})^{1 + 1/ n} + K

(D)

\frac{1}{n + 1} (1 + n x^{n})^{1 + 1/ n} + K

Answer:A

MathematicsEvaluation of Special Integral FormsJEE Advanced 2008Moderate

View in:English Hindi

Let $I = \int \frac{e ^{x}}{e ^{4 x} + e ^{2 x} + 1} d x, J = \int \frac{e ^{- x}}{e ^{- 4 x} + e ^{- 2 x} + 1} d x$ . Then, for an arbitrary constant $C$ , the value of $J - I$ equals

(A)

\frac{1}{2} lo g \frac{e ^{4 x} - e ^{2 x} + 1}{e ^{4 x} + e ^{2 x} + 1} + C

(B)

\frac{1}{2} lo g \frac{e ^{2 x} + e ^{x} + 1}{e ^{2 x} - e ^{x} + 1} + C

(C)

\frac{1}{2} lo g \frac{e ^{2 x} - e ^{x} + 1}{e ^{2 x} + e ^{x} + 1} + C

(D)

\frac{1}{2} lo g \frac{e ^{4 x} + e ^{2 x} + 1}{e ^{4 x} - e ^{2 x} + 1} + C

Answer:C

MathematicsIntegration by SubstitutionJEE Advanced 2012Moderate

View in:English Hindi

The integral $\int \frac{s e c ^{2} x}{( s e c x + t a n x ) ^{9/2}} d x$ equals (for some arbitrary constant $K$ )

(A)

\frac{1}{( s e c x + t a n x ) ^{11/2}} {\frac{1}{11} - \frac{1}{7} (sec x + tan x)^{2}} + K

(B)

\frac{1}{( s e c x + t a n x ) ^{11/2}} {\frac{1}{11} - \frac{1}{7} (sec x + tan x)^{2}} + K

(C)

\frac{1}{( s e c x + t a n x ) ^{11/2}} {\frac{1}{11} + \frac{1}{7} (sec x + tan x)^{2}} + K

(D)

\frac{1}{( s e c x + t a n x ) ^{11/2}} {\frac{1}{11} + \frac{1}{7} (sec x + tan x)^{2}} + K

Answer:C

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluate ∫x(1+xex)2(x+1)​dx

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Integrate ∫(x2+1)2(x+1)x3+3x+2​dx

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluate ∫sin−1(4x2+8x+13​2x+2​)dx

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For any natural number m, evaluate ∫(x3m+x2m+xm)(2x2m+3xm+6)1/mdx,x>0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of the integral ∫sin2x+sin4xcos3x+cos5x​dx is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫x32x4−2x2+1​x2−1​dx=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let F(x) be an indefinite integral of sin2x. STATEMENT-1: The function F(x) satisfies F(x+π)=F(x) because STATEMENT-2: sin2(x+π)=sin2x for all real x.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=(1+xn)1/nx​ for n≥2 and g(x)=(f∘f∘...∘f)(x) (f occurs n times). Then ∫xn−2g(x)dx equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let I=∫e4x+e2x+1ex​dx,J=∫e−4x+e−2x+1e−x​dx. Then, for an arbitrary constant C, the value of J−I equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The integral ∫(secx+tanx)9/2sec2x​dx equals (for some arbitrary constant K)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluate ∫x(1+xex)2(x+1)​dx

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Integrate ∫(x2+1)2(x+1)x3+3x+2​dx

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluate ∫sin−1(4x2+8x+13​2x+2​)dx

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For any natural number m, evaluate ∫(x3m+x2m+xm)(2x2m+3xm+6)1/mdx,x>0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of the integral ∫sin2x+sin4xcos3x+cos5x​dx is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫x32x4−2x2+1​x2−1​dx=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let F(x) be an indefinite integral of sin2x. STATEMENT-1: The function F(x) satisfies F(x+π)=F(x) because STATEMENT-2: sin2(x+π)=sin2x for all real x.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=(1+xn)1/nx​ for n≥2 and g(x)=(f∘f∘...∘f)(x) (f occurs n times). Then ∫xn−2g(x)dx equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let I=∫e4x+e2x+1ex​dx,J=∫e−4x+e−2x+1e−x​dx. Then, for an arbitrary constant C, the value of J−I equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The integral ∫(secx+tanx)9/2sec2x​dx equals (for some arbitrary constant K)

Evaluate $\int \frac{( x + 1 )}{x ( 1 + x e ^{x} ) ^{2}} d x$

Integrate $\int \frac{x ^{3} + 3 x + 2}{( x ^{2} + 1 ) ^{2} ( x + 1 )} d x$

Evaluate $\int sin^{- 1} (\frac{2 x + 2}{4 x ^{2} + 8 x + 13}) d x$

For any natural number $m$ , evaluate $\int (x^{3 m} + x^{2 m} + x^{m}) (2 x^{2 m} + 3 x^{m} + 6)^{1/ m} d x, x > 0$

The value of the integral $\int \frac{c o s ^{3} x + c o s ^{5} x}{s i n ^{2} x + s i n ^{4} x} d x$ is

$\int \frac{x ^{2} - 1}{x ^{3} 2 x ^{4} - 2 x ^{2} + 1} d x =$

Let $F (x)$ be an indefinite integral of $sin^{2} x$ . STATEMENT-1: The function $F (x)$ satisfies $F (x + π) = F (x)$ because STATEMENT-2: $sin^{2} (x + π) = sin^{2} x$ for all real $x$ .

Let $f (x) = \frac{x}{( 1 + x ^{n} ) ^{1/ n}}$ for $n \geq 2$ and $g (x) = (f \circ f \circ ... \circ f) (x)$ ( $f$ occurs $n$ times). Then $\int x^{n - 2} g (x) d x$ equals

Let $I = \int \frac{e ^{x}}{e ^{4 x} + e ^{2 x} + 1} d x, J = \int \frac{e ^{- x}}{e ^{- 4 x} + e ^{- 2 x} + 1} d x$ . Then, for an arbitrary constant $C$ , the value of $J - I$ equals

The integral $\int \frac{s e c ^{2} x}{( s e c x + t a n x ) ^{9/2}} d x$ equals (for some arbitrary constant $K$ )

Evaluate $\int \frac{( x + 1 )}{x ( 1 + x e ^{x} ) ^{2}} d x$

Integrate $\int \frac{x ^{3} + 3 x + 2}{( x ^{2} + 1 ) ^{2} ( x + 1 )} d x$

Evaluate $\int sin^{- 1} (\frac{2 x + 2}{4 x ^{2} + 8 x + 13}) d x$

For any natural number $m$ , evaluate $\int (x^{3 m} + x^{2 m} + x^{m}) (2 x^{2 m} + 3 x^{m} + 6)^{1/ m} d x, x > 0$

The value of the integral $\int \frac{c o s ^{3} x + c o s ^{5} x}{s i n ^{2} x + s i n ^{4} x} d x$ is

$\int \frac{x ^{2} - 1}{x ^{3} 2 x ^{4} - 2 x ^{2} + 1} d x =$

Let $F (x)$ be an indefinite integral of $sin^{2} x$ . STATEMENT-1: The function $F (x)$ satisfies $F (x + π) = F (x)$ because STATEMENT-2: $sin^{2} (x + π) = sin^{2} x$ for all real $x$ .

Let $f (x) = \frac{x}{( 1 + x ^{n} ) ^{1/ n}}$ for $n \geq 2$ and $g (x) = (f \circ f \circ ... \circ f) (x)$ ( $f$ occurs $n$ times). Then $\int x^{n - 2} g (x) d x$ equals

Let $I = \int \frac{e ^{x}}{e ^{4 x} + e ^{2 x} + 1} d x, J = \int \frac{e ^{- x}}{e ^{- 4 x} + e ^{- 2 x} + 1} d x$ . Then, for an arbitrary constant $C$ , the value of $J - I$ equals

The integral $\int \frac{s e c ^{2} x}{( s e c x + t a n x ) ^{9/2}} d x$ equals (for some arbitrary constant $K$ )