Question
Evaluate the following integrals:
$\int_{0}^\limits{1}\frac{1}{1+2\text{x}+2\text{x}^2+2\text{x}^3+\text{x}^4}\text{ dx}$
$\int_{0}^\limits{1}\frac{1}{1+2\text{x}+2\text{x}^2+2\text{x}^3+\text{x}^4}\text{ dx}$
✓
Answer
$\int_{0}^\limits{1}\frac{1}{1+2\text{x}+2\text{x}^2+2\text{x}^3+\text{x}^4}\text{ dx}$
$=\int_{0}^\limits{1}\frac{1}{\big(\text{x}^2+1\big)^2+2\text{x}\big(\text{x}^2+1\big)}\text{ dx}$
$=\int_{0}^\limits{1}\frac{1}{\big(\text{x}^2+1\big)\big(\text{x}^2+1+2\text{x}\big)}\text{ dx}$
$=\int_{0}^\limits{1}\frac{1}{(\text{x}^2+1)(\text{x}+1)}\text{ dx}$
Let $\frac{1}{(\text{x}+1)^2(\text{x}^2+1)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{B}}{(\text{x}+1)^2}+\frac{\text{Cx}+\text{D}}{\text{x}^2+1}$
$\Rightarrow1=\text{A}(\text{x}+1)(\text{x}^2+1)+\text{B}(\text{x}^2+1)+(\text{Cx}+\text{D})(\text{x}+1)^2$
Putting x = -1, we have
$1=2\text{B}$
$\Rightarrow\text{B}=\frac{1}{2}\ ...(\text{i})$
Putting x = 0, we have
$\text{A}+\text{B}+\text{C}=1\ ...(\text{ii})$
Equating co-efficient of x3 on both sides, we have
$\text{A}+\text{C}=0\ ...(\text{iii}) $
Equating co-efficient of x2 on both sides, we have
$\text{A}+\text{B}+2\text{C}+\text{D}=0\ ...(\text{iv})$
$\Rightarrow2\text{C}=-1$ [Using (i)]
$\Rightarrow\text{C}=-\frac{1}{2}$
$\therefore\ \text{A}=\frac{1}{2}$ [Using (iii)]
Putting $\text{A}=\frac{1}{2},\text{ B}=\frac{1}{2}$ and $\text{C}=-\frac{1}{2}$ in (iv), we have
$\text{D}=0$
$\therefore\ \int_{0}^\limits{1}\frac{1}{(\text{x}+1)^2(\text{x}^2+1)}\text{ dx}$
$=\int_{0}^\limits{1}\frac{\frac{1}{2}}{\text{x}+1}\text{ dx}+\int_{0}^\limits{1}\frac{\frac{1}{2}}{(\text{x}+1)^2}\text{ dx}+\int_{0}^\limits{1}\frac{-\frac{1}{2}\text{x}}{\text{x}^2+1}$
$=\Big[\frac{1}{2}\log(\text{x}+1)\Big]^1_0+\Big[\frac{1}{2}\times\Big(-\frac{1}{\text{x}+1}\Big)\Big]^1_0-\frac{1}{4}\int_{0}^\limits{1}\frac{2\text{x}}{\text{x}^2+1}\text{ dx}$
$=\frac{1}{2}\big(\log2-\log1\big)-\frac{1}{2}\Big(\frac{1}{2}-1\Big)-\Big[\frac{1}{4}\log(\text{x}^2+1)\Big]^1_0$
$=\frac{1}{2}\log2+\frac{1}{4}-\frac{1}{4}\big(\log2-\log1\big)$ $(\log1=0)$
$=\frac{1}{2}\log2+\frac{1}{4}\log\text{e}-\frac{1}{4}\log2$
$=\frac{1}{4}\log2+\frac{1}{4}\log_\text{e}$
$=\frac{1}{4}\big(\log2+\log_\text{e}\big)$
$=\frac{1}{4}\log(2\text{e})$
$=\int_{0}^\limits{1}\frac{1}{\big(\text{x}^2+1\big)^2+2\text{x}\big(\text{x}^2+1\big)}\text{ dx}$
$=\int_{0}^\limits{1}\frac{1}{\big(\text{x}^2+1\big)\big(\text{x}^2+1+2\text{x}\big)}\text{ dx}$
$=\int_{0}^\limits{1}\frac{1}{(\text{x}^2+1)(\text{x}+1)}\text{ dx}$
Let $\frac{1}{(\text{x}+1)^2(\text{x}^2+1)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{B}}{(\text{x}+1)^2}+\frac{\text{Cx}+\text{D}}{\text{x}^2+1}$
$\Rightarrow1=\text{A}(\text{x}+1)(\text{x}^2+1)+\text{B}(\text{x}^2+1)+(\text{Cx}+\text{D})(\text{x}+1)^2$
Putting x = -1, we have
$1=2\text{B}$
$\Rightarrow\text{B}=\frac{1}{2}\ ...(\text{i})$
Putting x = 0, we have
$\text{A}+\text{B}+\text{C}=1\ ...(\text{ii})$
Equating co-efficient of x3 on both sides, we have
$\text{A}+\text{C}=0\ ...(\text{iii}) $
Equating co-efficient of x2 on both sides, we have
$\text{A}+\text{B}+2\text{C}+\text{D}=0\ ...(\text{iv})$
$\Rightarrow2\text{C}=-1$ [Using (i)]
$\Rightarrow\text{C}=-\frac{1}{2}$
$\therefore\ \text{A}=\frac{1}{2}$ [Using (iii)]
Putting $\text{A}=\frac{1}{2},\text{ B}=\frac{1}{2}$ and $\text{C}=-\frac{1}{2}$ in (iv), we have
$\text{D}=0$
$\therefore\ \int_{0}^\limits{1}\frac{1}{(\text{x}+1)^2(\text{x}^2+1)}\text{ dx}$
$=\int_{0}^\limits{1}\frac{\frac{1}{2}}{\text{x}+1}\text{ dx}+\int_{0}^\limits{1}\frac{\frac{1}{2}}{(\text{x}+1)^2}\text{ dx}+\int_{0}^\limits{1}\frac{-\frac{1}{2}\text{x}}{\text{x}^2+1}$
$=\Big[\frac{1}{2}\log(\text{x}+1)\Big]^1_0+\Big[\frac{1}{2}\times\Big(-\frac{1}{\text{x}+1}\Big)\Big]^1_0-\frac{1}{4}\int_{0}^\limits{1}\frac{2\text{x}}{\text{x}^2+1}\text{ dx}$
$=\frac{1}{2}\big(\log2-\log1\big)-\frac{1}{2}\Big(\frac{1}{2}-1\Big)-\Big[\frac{1}{4}\log(\text{x}^2+1)\Big]^1_0$
$=\frac{1}{2}\log2+\frac{1}{4}-\frac{1}{4}\big(\log2-\log1\big)$ $(\log1=0)$
$=\frac{1}{2}\log2+\frac{1}{4}\log\text{e}-\frac{1}{4}\log2$
$=\frac{1}{4}\log2+\frac{1}{4}\log_\text{e}$
$=\frac{1}{4}\big(\log2+\log_\text{e}\big)$
$=\frac{1}{4}\log(2\text{e})$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin 3 times and notes the number of heads. If she gets 1, 2, 3 or 4 she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=\text{x}^3$
→$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=\text{x}^3$
Find the matrix A such that
$\begin{bmatrix}1&1\\0&1\end{bmatrix}\text{A}=\begin{bmatrix}3&3&5\\1&0&1\end{bmatrix}$
→$\begin{bmatrix}1&1\\0&1\end{bmatrix}\text{A}=\begin{bmatrix}3&3&5\\1&0&1\end{bmatrix}$
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=(\sin\text{x})^\text{x}+\sin^{-1}\sqrt{\text{x}}$
→$\text{y}=(\sin\text{x})^\text{x}+\sin^{-1}\sqrt{\text{x}}$
Evaluate the following integrals:
$\int\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$
→$\int\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$
Find the intervals in which the following functions are increasing or decreasing.
f(x) = x4 - 4x3 - 4x2 + 15
f(x) = x4 - 4x3 - 4x2 + 15
Let A = {1, 2, 3, ... 9} and R be the relation in A × A defined by (a, b)R(c, d) if a + d = b + c for (a, b), (c, d) in A × A. Prove that R is an equivalence relation and also obtain the equivalent class [(2, 5)].
A manufacturer considers that men and women workers are equally efficient and so he pays them at the same rate. He has 30 and 17 units of workers (male and female) and capital respectively, which he uses to produce two types of goods A and B. To produce one unit of A, 2 workers and 3 units of capital are required while 3 workers and 1 unit of capital is required to produce one unit of B. If A and B are priced at Rs. 100 and Rs. 120 per unit respectively, how should he use his resources to maximise the total revenue? Form the above as an LPP and solve graphically. Do you agree with this view of the manufacturer that men and women workers are equally efficient and so should be paid at the same rate?
The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting at least twice?
Find the values of p and q, for which
![]( "Click and drag to move")
is continuous at![]()
![]( "Click and drag to move")
is continuous at