Question:
$ \int{(x+1){{(x+2)}^{7}}}(x+3)dx $ is equal to
$ \int{(x+1){{(x+2)}^{7}}}(x+3)dx $ is equal to
Updated On: Jun 6, 2022
- $ \frac{{{(x+2)}^{10}}}{10}-\frac{{{(x+2)}^{8}}}{8}+C $
- $ \frac{{{(x+1)}^{2}}}{2}-\frac{{{(x+2)}^{8}}}{8}-\frac{{{(x+3)}^{2}}}{2}+C $
- $ \frac{{{(x+2)}^{10}}}{10}+C $
- $ \frac{{{(x+1)}^{2}}}{2}+\frac{{{(x+2)}^{8}}}{8}+\frac{{{(x+3)}^{2}}}{2}+C $
Hide Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Let $ I=\int{(x+1){{(x+2)}^{7}}(x+3)}dx $
Putting $ x+2=t $
$ \Rightarrow $ $ dx=dt $
Also, $ x+1=t-1 $ and $ x+3=t+1 $
$ \therefore $ $ I=\int{(t-1){{t}^{7}}}(t+1)dt $
$=\int{({{t}^{2}}-1)}{{t}^{7}}dt $
$=\int{{{t}^{9}}}dt-\int{{{t}^{7}}}dt $
$=\frac{{{t}^{10}}}{10}-\frac{{{t}^{8}}}{8}+c $
$=\frac{{{(x+2)}^{10}}}{10}-\frac{{{(x+2)}^{8}}}{8}+c $
Putting $ x+2=t $
$ \Rightarrow $ $ dx=dt $
Also, $ x+1=t-1 $ and $ x+3=t+1 $
$ \therefore $ $ I=\int{(t-1){{t}^{7}}}(t+1)dt $
$=\int{({{t}^{2}}-1)}{{t}^{7}}dt $
$=\int{{{t}^{9}}}dt-\int{{{t}^{7}}}dt $
$=\frac{{{t}^{10}}}{10}-\frac{{{t}^{8}}}{8}+c $
$=\frac{{{(x+2)}^{10}}}{10}-\frac{{{(x+2)}^{8}}}{8}+c $
Was this answer helpful?
0
0
Top Questions on Integrals of Some Particular Functions
- If \(\int \frac{dx}{(x+1)(x-2)(x-3)}=\frac{1}{k}log_e\left \{ \frac{|x-3|^3|x+1|}{(x-2)^4}\right \}+c\), then the value of k is
- WBJEE - 2023
- Mathematics
- Integrals of Some Particular Functions
- Let $f(x)=\int \frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)} d x$ If $f(3)=\frac{1}{2}\left(\log _e 5-\log _e 6\right)$, then $f(4)$ is equal to
- JEE Main - 2023
- Mathematics
- Integrals of Some Particular Functions
- If $\int \sqrt{\sec 2 x-1} d x=\alpha \log _e\left|\cos 2 x+\beta+\sqrt{\cos 2 x\left(1+\cos \frac{1}{\beta} x\right)}\right|+$ constant, then $\beta-\alpha$ is equal to_______
- JEE Main - 2023
- Mathematics
- Integrals of Some Particular Functions
- If $\int \limits_0^\pi \frac{5^{\cos x}\left(1+\cos x \cos 3 x+\cos ^2 x+\cos ^3 x \cos 3 x\right) d x}{1+5^{\cos x}}=\frac{ k \pi}{16}$, then $k$ is equal to
- JEE Main - 2023
- Mathematics
- Integrals of Some Particular Functions
- The integral $16 \int\limits_1^2 \frac{d x}{x^3\left(x^2+2\right)^2}$ is equal to
- JEE Main - 2023
- Mathematics
- Integrals of Some Particular Functions
View More Questions
Questions Asked in KEAM exam
- A Spherical ball is subjected to a pressure of 100 atmosphere. If the bulk modulus of the ball is 1011 N/m2,then change in the volume is:
- KEAM - 2023
- mechanical properties of solids
- Two identical solid spheres,each of radius 10cm,are kept in contact. If the mass of inertia of this system about the tangent passing through the point of contact is 0. Then mass of each sphere is:
- KEAM - 2023
- System of Particles & Rotational Motion
- The value of a(≠0) for which the equation \(\frac{1}{2}(x-2)^2+1=\sin(\frac{a}{x})\) holds is/are
- KEAM - 2023
- Trigonometric Functions
- \(∫xlog(1+x^2)dx=\)?
- KEAM - 2023
- Integration by Parts
- A uniform thin rod of mass 3kg has a length of 1m. If a point mass of 1kg is attached to it at a distance of 40cm from its center,the center of mass shifts by a distance of:
- KEAM - 2023
- System of Particles & Rotational Motion
View More Questions
Concepts Used:
Integrals of Some Particular Functions
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
Integrals of Some Particular Functions:
- ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
- ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
- ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
- ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
- ∫1/√(a2 – x2) dx = sin-1(x/a) + C
- ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C
These are tabulated below along with the meaning of each part.
